🔬 Model Lab

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
---
### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
---
### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
---
### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
---
### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
---
### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

---

**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

---

**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

---

**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

---

**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

---

**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

---

**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

---

**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

---

**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

---

**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

---

**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

---

**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

---

**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B

I'll solve each problem step-by-step.

## Problem 1:

**What is being asked:** Find the maximum number of additional license plates by adding 2 letters to the three sets.

**Current setup:**
- Set 1: 5 letters → 5 choices
- Set 2: 3 letters → 3 choices
- Set 3: 4 letters → 4 choices
- Current plates: 5 × 3 × 4 = 60

**Strategy:** Add letters to maximize the product. Let's test options:

1) Add both to Set 1: (5+2) × 3 × 4 = 7 × 3 × 4 = 84 plates → 84 - 60 = **24 additional**
2) Add both to Set 2: 5 × (3+2) × 4 = 5 × 5 × 4 = 100 plates → 100 - 60 = **40 additional**
3) Add both to Set 3: 5 × 3 × (4+2) = 5 × 3 × 6 = 90 plates → 90 - 60 = **30 additional**
4) Add 1 to Set 1, 1 to Set 2: 6 × 4 × 4 = 96 plates → 96 - 60 = **36 additional**
5) Add 1 to Set 1, 1 to Set 3: 6 × 3 × 5 = 90 plates → 90 - 60 = **30 additional**
6) Add 1 to Set 2, 1 to Set 3: 5 × 4 × 5 = 100 plates → 100 - 60 = **40 additional**

Maximum is **40 additional plates** (achieved by adding both to Set 2, or 1 to Set 2 and 1 to Set 3).

ANSWER 1: D

## Problem 2:

**What is being asked:** Find average miles per gallon for the entire trip.

**Analysis:**
- Start: 56,200 miles, filled with 6 gallons
- First refill: 56,560 miles, filled with 12 gallons
- Second refill: 57,060 miles, filled with 20 gallons

The gas used between refills equals the amount needed to refill:
- From start to first refill: 12 gallons used, distance = 56,560 - 56,200 = 360 miles
- From first refill to second refill: 20 gallons used, distance = 57,060 - 56,560 = 500 miles

Total distance: 360 + 500 = 860 miles
Total gas used: 12 + 20 = 32 gallons

Average MPG: 860 ÷ 32 = 26.875 ≈ **26.9**

ANSWER 2: D

## Problem 3:

**What is being asked:** Find the sum of the two lowest test scores.

**Given:** 5 scores, mean = 90, median = 91, mode = 94

**Setup:** Sum of all 5 scores = 5 × 90 = 450

Let the scores in order be: a ≤ b ≤ c ≤ d ≤ e

- Median is 91: c = 91
- Mode is 94: 94 appears more than any other value
- Since we need a mode of 94, at least two scores must be 94

Since c = 91 and we need two 94s, they must be d and e.
So: a ≤ b ≤ 91 ≤ 94 ≤ 94

Sum: a + b + 91 + 94 + 94 = 450
a + b = 450 - 279 = **171**

ANSWER 3: B

## Problem 4:

**What is being asked:** How many turns until Alice and Bob meet on the same point?

**Setup:** 12 points on circle, both start at point 12.
- Alice moves 5 points clockwise per turn
- Bob moves 9 points counterclockwise per turn

After n turns:
- Alice is at position: 12 + 5n (mod 12)
- Bob is at position: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
- 5n ≡ -9n (mod 12)
- 14n ≡ 0 (mod 12)
- 14n = 12k for some integer k

Since 14n ≡ 2n (mod 12), we need: 2n ≡ 0 (mod 12)
- n ≡ 0 (mod 6)

Smallest positive n = **6**

ANSWER 4: A

## Problem 5:

**What is being asked:** Find the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Calculation:**
- 2!! = 2 (units digit: 2)
- 4!! = 2 × 4 = 8 (units digit: 8)
- 6!! = 2 × 4 × 6 = 48 (units digit: 8)
- 8!! = 2 × 4 × 6 × 8 = 384 (units digit: 4)
- 10!! = 2 × 4 × 6 × 8 × 10 = 3,840 (units digit: 0)

For all n ≥ 10, n!! includes the factors 2, 4, 6, 8, and 10, giving at least two factors of 5 and multiple factors of 2, so units digit is 0.

Sum of units digits: 2 + 8 + 8 + 4 + 0 + 0 + ... = 22 (units digit: **2**)

ANSWER 5: B

## Problem 6:

**What is being asked:** Count whole numbers from 1 to 46 divisible by 3 or 5.

**Using inclusion-exclusion:**
- Divisible by 3: ⌊46/3⌋ = 15
- Divisible by 5: ⌊46/5⌋ = 9
- Divisible by both (15): ⌊46/15⌋ = 3

Total: 15 + 9 - 3 = **21**

ANSWER 6: B

## Problem 7:

**What is being asked:** Find the number of guppies in the 4th tank.

**Setup:** Let tank 1 have x guppies:
- Tank 1: x
- Tank 2: x + 1
- Tank 3: x + 1 + 2 = x + 3
- Tank 4: x + 3 + 3 = x + 6

Total: x + (x+1) + (x+3) + (x+6) = 90
- 4x + 10 = 90
- 4x = 80
- x = 20

Tank 4: 20 + 6 = **26**

ANSWER 7: E

## Problem 8:

**What is being asked:** Find the percentage increase in area when length increases by 20% and width by 50%.

**Setup:**
- Original area: A = LW
- New length: 1.20L
- New width: 1.50W
- New area: (1.20L)(1.50W) = 1.80LW

Percentage increase: (1.80 - 1.00) × 100% = **80%**

ANSWER 8: D

## Problem 9:

**What is being asked:** Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:** a⁴ - b⁴ = (a² - b²)(a² + b²) = (a-b)(a+b)(a² + b²)

13⁴ - 11⁴ = (13-11)(13+11)(13² + 11²) = 2 × 24 × (169 + 121) = 2 × 24 × 290

- 2 = 2¹
- 24 = 2³ × 3
- 290 = 2 × 145

Total: 2^(1+3+1) × (3 × 145) = 2⁵ × 435 = **32** is the largest power of 2

ANSWER 9: C

## Problem 10:

**What is being asked:** Find the average of the four distances from the lemming's final position to each side of the square.

**Setup:** Place the square with corners at (0,0), (10,0), (10,10), (0,10). Lemming starts at (0,0).

Diagonal toward opposite corner is to (10,10), direction: (1,1)/√2

After 6.2 meters: position = (6.2/√2, 6.2/√2) ≈ (4.384, 4.384)

Turns 90° right. The perpendicular direction is (1,-1)/√2.

After 2 more meters: position ≈ (4.384 + 2/√2, 4.384 - 2/√2) ≈ (5.8, 3.0)

Distances to four sides:
- To x=0: 5.8
- To x=10: 4.2
- To y=0: 3.0
- To y=10: 7.0

Average: (5.8 + 4.2 + 3.0 + 7.0)/4 = 20/4 = **5**

ANSWER 10: C

## Problem 11:

**What is being asked:** Find the largest of 25 consecutive even integers with sum 10,000.

**Setup:** Let the integers be 2n, 2n+2, 2n+4, ..., 2n+48

Sum = 25(2n) + 2(0+2+4+...+48) = 50n + 2(0+2+...+48)

The sum 0+2+4+...+48 = 2(0+1+2+...+24) = 2 × (24×25/2) = 600

So: 50n + 1200 = 10,000
- 50n = 8,800
- n = 176

Largest: 2(176) + 48 = 352 + 48 = **400**

Wait, let me recalculate. If middle term is 2m+24, then:
Sum = 25 × (middle term) = 25 × (2m+24) = 10,000
- 2m + 24 = 400
- m = 188

Largest: 188 + 24 = **212**, no...

Actually: 25 consecutive even integers centered at the 13th term.
If 13th term is k, sum = 25k = 10,000, so k = 400.
Largest = 400 + 12×2 = **424**

ANSWER 11: E

## Problem 12:

**What is being asked:** Find the actual time when the car cloc

Problem 1:

Let the original number of plates be found by multiplying the choices for each letter position:

- 1st letter: 5 choices
- 2nd letter: 3 choices
- 3rd letter: 4 choices

So originally there are
\[
5\cdot 3\cdot 4=60
\]
plates.

Now two new letters are added. To get the **largest possible** increase, we should add both letters to the set with the most choices, because that gives the biggest multiplication effect.

Let the set sizes be \(5,3,4\). The biggest set is the first one, with 5 letters. Add 2 letters there, making it 7 choices.

New number of plates:
\[
7\cdot 3\cdot 4=84
\]

So the number of **additional** plates is
\[
84-60=24
\]

If we split the two letters between different sets, the increase is smaller:
- Add one to the 5-letter set and one to the 4-letter set:
  \[
  6\cdot 3\cdot 5=90
  \]
  Wait, this is actually an increase of \(30\), which is larger than 24.
- Add one to the 5-letter set and one to the 3-letter set:
  \[
  6\cdot 4\cdot 4=96
  \]
  increase \(36\).
- Add both to the 4-letter set:
  \[
  5\cdot 3\cdot 6=90
  \]
  increase \(30\).
- Add both to the 3-letter set:
  \[
  5\cdot 5\cdot 4=100
  \]
  increase \(40\).

The best is to add one letter to the 5-letter set and one to the 3-letter set, giving:
\[
6\cdot 4\cdot 4=96
\]
So the number of additional plates is
\[
96-60=36
\]

ANSWER 1: C

Problem 2:

The car’s average miles per gallon is

\[
\frac{\text{total miles driven}}{\text{total gallons used}}
\]

First find the miles driven:
- From 56,200 to 56,560: \(360\) miles
- From 56,560 to 57,060: \(500\) miles

So total distance is
\[
360+500=860 \text{ miles}
\]

Total gasoline used is
\[
6+12+20=38 \text{ gallons}
\]

Thus the average mpg is
\[
\frac{860}{38}\approx 22.6316
\]
Rounded to the nearest tenth:
\[
22.6
\]

ANSWER 2: B

Problem 3:

Let the five scores in order be
\[
a\le b\le c\le d\le e
\]
We are told:

- Mean is 90, so sum is
  \[
  a+b+c+d+e=450
  \]
- Median is 91, so
  \[
  c=91
  \]
- Mode is 94, so 94 occurs more often than any other score.

Since there are only five scores and the median is 91, the only way for 94 to be the mode is for the last two scores to be 94:
\[
d=e=94
\]
Then the sum becomes
\[
a+b+91+94+94=450
\]
So
\[
a+b=450-279=171
\]

ANSWER 3: B

Problem 4:

Alice starts at 12 and moves 5 points clockwise each turn.
Bob starts at 12 and moves 9 points counterclockwise each turn.

Instead of tracking both separately, look at their relative movement.

From Alice’s point of view, Bob moves:
- 9 points counterclockwise = 3 points clockwise, since the circle has 12 points.

So each turn:
- Alice moves 5 clockwise
- Bob moves 3 clockwise

Their separation changes by
\[
5-3=2
\]
points clockwise each turn.

They start together at point 12, and they meet again when the separation is a multiple of 12. Since each turn changes the separation by 2, we need
\[
2t\equiv 0 \pmod{12}
\]
The smallest positive \(t\) is
\[
t=6
\]

Check:
- After 6 turns, separation is \(12\equiv 0\), so they meet.

ANSWER 4: A

Problem 5:

We need the units digit of
\[
2!!+4!!+6!!+\cdots+2022!!
\]

For even \(n\ge 2\), \(n!!\) is the product of all even numbers from 2 to \(n\). Every term except \(2!!\) contains a factor of 10? Let’s inspect the units digit.

- \(2!!=2\)
- \(4!!=2\cdot 4=8\)
- \(6!!=2\cdot 4\cdot 6=48\), units digit 8
- \(8!!\) ends in 8 as well
- Once the product includes both 2 and 5, it ends in 0. But even double factorials contain no 5, so we should be careful.

Let’s compute modulo 10:
For \(n\ge 4\), \(n!!\) includes a factor 2 and a factor 4, so it is divisible by 8. In fact:
- \(4!!\equiv 8 \pmod{10}\)
- \(6!!=48\equiv 8\)
- \(8!!=384\equiv 4\)? Let's find the pattern more carefully.

Better: write
\[
n!!=2\cdot 4\cdot 6\cdots n=2^{k}(1\cdot 2\cdot 3\cdots k)=2^k k!
\]
where \(n=2k\).

For large \(k\), the units digit repeats. But a simpler observation: for \(k\ge 5\), \(k!\) is divisible by 10, so \(2^k k!\) ends in 0. Since \(2022=2\cdot 1011\), all terms from \(10!!\) onward end in 0.

So only the first few terms matter:
\[
2!!=2,\quad 4!!=8,\quad 6!!=48,\quad 8!!=384,\quad 10!! \text{ and beyond end in }0
\]

Units digits:
\[
2,8,8,4
\]
Sum:
\[
2+8+8+4=22
\]
Units digit is
\[
2
\]

ANSWER 5: B

Problem 6:

Count numbers from 1 through 46 divisible by 3 or 5.

Multiples of 3:
\[
\left\lfloor \frac{46}{3}\right\rfloor=15
\]
Multiples of 5:
\[
\left\lfloor \frac{46}{5}\right\rfloor=9
\]
Multiples of both 3 and 5, i.e. 15:
\[
\left\lfloor \frac{46}{15}\right\rfloor=3
\]

By inclusion-exclusion:
\[
15+9-3=21
\]

ANSWER 6: B

Problem 7:

Let the first tank have \(x\) guppies.

Then the tanks have:
- 1st: \(x\)
- 2nd: \(x+1\)
- 3rd: \(x+3\)
- 4th: \(x+6\)

Their total is 90:
\[
x+(x+1)+(x+3)+(x+6)=90
\]
\[
4x+10=90
\]
\[
4x=80
\]
\[
x=20
\]

So the 4th tank has
\[
x+6=26
\]

ANSWER 7: E

Problem 8:

If length increases by 20%, new length is \(1.2L\).

If width increases by 50%, new width is \(1.5W\).

So new area is
\[
1.2L\cdot 1.5W=1.8LW
\]
That is 180% of the original area, so the increase is
\[
180\%-100\%=80\%
\]

ANSWER 8: D

Problem 9:

We need the largest power of 2 dividing
\[
13^4-11^4
\]

Use difference of squares:
\[
13^4-11^4=(13^2-11^2)(13^2+11^2)
\]
\[
=(169-121)(169+121)=48\cdot 290
\]

Now factor powers of 2:
\[
48=16\cdot 3=2^4\cdot 3
\]
\[
290=2\cdot 145
\]
So the product has
\[
2^4\cdot 2=2^5
\]
as its highest power of 2, unless there is another factor of 2 hidden in 145, but 145 is odd.

Thus the largest power of 2 dividing the expression is
\[
2^5=32
\]

ANSWER 9: C

Problem 10:

The square has side length 10.

The lemming starts at a corner, then runs 6.2 meters along a diagonal toward the opposite corner. Since the diagonal of a 10-by-10 square has length \(10\sqrt2\), moving 6.2 meters along it puts the lemming somewhere inside the square. Then it turns right and runs 2 more meters.

We are asked for the average of the shortest distances from the lemming to the four sides.

A point inside a square has

Problem 1:  
We form license plates with 3 positions:
- 1st letter from {C, H, L, P, R} → 5 choices  
- 2nd letter from {A, I, O} → 3 choices  
- 3rd letter from {D, M, N, T} → 4 choices  

Total plates originally: \(5\cdot 3\cdot 4=60\).

They add two new letters. We must decide where to add them to maximize the new total number of plates.  

Adding two letters can happen in two ways:
1) Add both to one set  
2) Add one to two different sets  

### Case 1: Add both to the same set
- If add both to the 1st set: \((5+2)\cdot 3\cdot 4 = 7\cdot 3\cdot 4=84\)  
  Additional: \(84-60=24\)
- If add both to the 2nd set: \(5\cdot (3+2)\cdot 4 = 5\cdot 5\cdot 4=100\)  
  Additional: \(100-60=40\)
- If add both to the 3rd set: \(5\cdot 3\cdot (4+2)=5\cdot 3\cdot 6=90\)  
  Additional: \(90-60=30\)

Best so far: 40 additional.

### Case 2: Add one to each of two sets
Try all distributions:

- Add to 1st and 2nd: \((5+1)(3+1)4=6\cdot 4\cdot 4=96\)  
  Additional: \(96-60=36\)
- Add to 1st and 3rd: \((5+1)3(4+1)=6\cdot 3\cdot 5=90\)  
  Additional: \(90-60=30\)
- Add to 2nd and 3rd: \(5(3+1)(4+1)=5\cdot 4\cdot 5=100\)  
  Additional: \(100-60=40\)

Maximum additional is \(40\).

ANSWER 1: D  

---

Problem 2:  
We read:
- Start odometer: 56,200  
- First fill: 6 gallons (at 56,200)  
- Second fill: 12 gallons at odometer 56,560  
- Third fill: 20 gallons at odometer 57,060  
We need average miles-per-gallon over the whole trip, i.e.  
\[
\text{miles-per-gallon}=\frac{\text{total miles driven}}{\text{total gallons used}}
\]

### Total miles driven
From 56,200 to 56,560: \(56,560-56,200=360\) miles  
From 56,560 to 57,060: \(57,060-56,560=500\) miles  
Total: \(360+500=860\) miles

### Total gallons used
\(6+12+20=38\) gallons

Average MPG:
\[
\frac{860}{38}=22.6315\ldots
\]
To the nearest tenth: \(22.6\).

ANSWER 2: B  

---

Problem 3:  
We have 5 scores with:
- Mean = 90 ⇒ total sum \(=5\cdot 90=450\)
- Median = 91 ⇒ the 3rd score (middle one) is 91  
- Mode = 94 ⇒ 94 appears most often

Let the sorted scores be:
\[
a \le b \le 91 \le d \le e
\]
But we need mode 94. Since 91 is the median, 94 must occur among the 4th and/or 5th positions.

The only way for 94 to be the mode (most frequent value) with 5 numbers is:
- 94 occurs twice (or more). The simplest consistent pattern is:
\[
a,\; b,\; 91,\; 94,\; 94
\]
Then 94 is the mode.

Now sum:
\[
a+b+91+94+94=450
\]
\[
a+b + 279 = 450
\]
\[
a+b = 171
\]

So the sum of the two lowest test scores is 171.

ANSWER 3: B  

---

Problem 4:  
Circle points numbered 1 to 12.  
Both start at point 12.

- Alice moves 5 points clockwise each turn.
- Bob moves 9 points counterclockwise each turn.
They stop when they land on the same point after a whole number of turns.

Work modulo 12.

Let’s track positions after \(t\) turns.

Alice (clockwise +5 each turn):
\[
A_t \equiv 12 + 5t \pmod{12}
\]
Since 12 ≡ 0 mod 12:
\[
A_t \equiv 5t \pmod{12}
\]

Bob moves counterclockwise 9 points, which is the same as clockwise \(-9\equiv +3\) (mod 12) each turn. Starting at 12 ≡ 0:
\[
B_t \equiv 3t \pmod{12}
\]

They meet when:
\[
5t \equiv 3t \pmod{12}
\]
\[
2t \equiv 0 \pmod{12}
\]
So \(12\mid 2t\), meaning \(6\mid t\).

Smallest positive \(t\) is \(t=6\).  

So it takes 6 turns.

ANSWER 4: A  

---

Problem 5:  
We want the units digit of:
\[
2!! + 4!! + 6!! + \cdots + 2022!!
\]
Where \(n!!\) is product of even numbers from 2 to \(n\).

Compute units digits patterns:
- \(2!! = 2\) → units digit 2
- \(4!!=2\cdot 4=8\) → units digit 8
- \(6!!=2\cdot 4\cdot 6\) ends in units digit \(0\) because it includes factor 10? Let’s check carefully:  
  \(2\cdot 4\cdot 6=48\), units digit 8. So not 0 yet.
Let’s look more systematically.

For any even \(n \ge 10\), the product \(2\cdot 4\cdot 6\cdots n\) includes a factor 10, 20, 30, etc., so it has a factor of 10 ⇒ units digit 0.

So:
- Only \(2!!, 4!!, 6!!, 8!!\) might contribute nonzero units digits.
- For \(n \ge 10\), \(n!!\equiv 0 \pmod{10}\).

Now compute:
\[
2!!=2
\]
\[
4!!=2\cdot 4=8
\]
\[
6!!=2\cdot 4\cdot 6=48 \Rightarrow 8
\]
\[
8!!=2\cdot 4\cdot 6\cdot 8 = 48\cdot 8=384 \Rightarrow 4
\]

Sum units digit:
\[
2+8+8+4=22
\]
Units digit of 22 is 2.

ANSWER 5: B  

---

Problem 6:  
Count whole numbers from 1 to 46 divisible by 3 or 5 (or both).

Use inclusion-exclusion.

Multiples of 3 up to 46:
\[
\left\lfloor \frac{46}{3}\right\rfloor = 15
\]
Multiples of 5 up to 46:
\[
\left\lfloor \frac{46}{5}\right\rfloor = 9
\]
Multiples of both (lcm of 3 and 5 is 15):
\[
\left\lfloor \frac{46}{15}\right\rfloor = 3
\]

Total:
\[
15+9-3=21
\]

ANSWER 6: B  

---

Problem 7:  
Let the number of guppies in tanks 1–4 be \(T_1,T_2,T_3,T_4\).

Given:
- \(T_2 = T_1 + 1\)
- \(T_3 = T_2 + 2\)
- \(T_4 = T_3 + 3\)

Total guppies:
\[
T_1+T_2+T_3+T_4=90
\]

Express in terms of \(T_1\):
\[
T_2=T_1+1
\]
\[
T_3=T_1+1+2=T_1+3
\]
\[
T_4=T_1+3+3=T_1+6
\]

Sum:
\[
T_1+(T_1+1)+(T_1+3)+(T_1+6)=90
\]
\[
4T_1+10=90
\]
\[
4T_1=80
\]
\[
T_1=20
\]
So
\[
T_4 = T_1+6=26
\]

ANSWER 7: E  

---

Problem 8:  
Area of rectangle is \(A = lw\).

- Length increased by 20% ⇒ new length \(=1.2l\)
- Width increased by 50% ⇒ new width \(=1.5w\)

New area:
\[
A' = (1.2l)(1.5w)=1.8lw=1.8A
\]
So area increases by:
\[
1.8A - A = 0.8A
\]
That is an 80% increase.

ANSWER 8: D  

---

Problem 9:  
We need the largest power of 2 dividing:
\[
13^4 - 11^4
\]

Factor using difference of fourth powers:
\[
13^4 - 11^4 = (13^2-11^2)(13^2+11^2)
\]
Compute:
\[
13^2=169,\quad 11^2=121
\]
\[
13^2-11^2 = 169-121=48
\]
\[
13^2+11^2 = 169+121=290
\]
So:
\[
13^4-11^4 = 48\cdot 290
\]

Now find highest power of 2 in this product.
\[
48=16\cdot 3 = 2^4\cdot 3
\]
\[
290=2\cdot 145 = 2^1\cdot 145
\]
Total power:
\[
2^{4+1}=2^5=32
\]

So the largest power of 2 divisor is 32.

ANSWER 9: C  

---

Problem 10:  
Square side length 10. Place coordinates:
- Start at corner \(A=(0,0)\)
- Opposite corner is \(C=(10,10)\)

The lemming runs 6.2 meters along a diagonal toward the opposite corner.  
So it goes from \((0,

Problem 1:
The original number of license plates is $5 \times 3 \times 4 = 60$. Let the sizes of the sets be $n_1=5, n_2=3, n_3=4$. Adding two letters means we increase the product. If we add both to set $i$, the new count is $(n_i+2) \times n_j \times n_k$. If we add one to set $i$ and one to set $j$, the new count is $(n_i+1) \times (n_j+1) \times n_k$.
- Add 2 to $n_1$: $7 \times 3 \times 4 = 84$ (Increase 24)
- Add 2 to $n_2$: $5 \times 5 \times 4 = 100$ (Increase 40)
- Add 2 to $n_3$: $5 \times 3 \times 6 = 90$ (Increase 30)
- Add 1 to $n_1, n_2$: $6 \times 4 \times 4 = 96$ (Increase 36)
- Add 1 to $n_1, n_3$: $6 \times 3 \times 5 = 90$ (Increase 30)
- Add 1 to $n_2, n_3$: $5 \times 4 \times 5 = 100$ (Increase 40)
The largest increase is 40.
ANSWER 1: D

Problem 2:
Total miles = $57,060 - 56,200 = 860$ miles. Total gas = $6 + 12 + 20 = 38$ gallons. Average = $860 / 38 \approx 22.63$. Rounded to the nearest tenth, this is 22.6.
ANSWER 2: B

Problem 3:
Let the scores be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. Mean = 90, so sum = 450. Median $x_3 = 91$. Mode = 94. Since 94 is the mode and there are 5 scores, at least two scores must be 94. Since $x_3=91$, the 94s must be $x_4$ and $x_5$. So $x_1 + x_2 + 91 + 94 + 94 = 450 \implies x_1 + x_2 + 279 = 450 \implies x_1 + x_2 = 171$.
ANSWER 3: B

Problem 4:
Alice starts at 12, moves +5 mod 12. Bob starts at 12, moves -9 mod 12. After $n$ turns, Alice is at $12 + 5n \pmod{12}$ and Bob is at $12 - 9n \pmod{12}$. They meet when $5n \equiv -9n \pmod{12} \implies 14n \equiv 0 \pmod{12} \implies 2n \equiv 0 \pmod{12}$. The smallest positive $n$ is 6.
ANSWER 4: A

Problem 5:
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$, $10!! = 3840$. For $n \ge 10$, $n!!$ ends in 0. Sum = $2 + 8 + 48 + 384 + 0 + \dots = 442$. The units digit is 2.
ANSWER 5: B

Problem 6:
Divisible by 3: $\lfloor 46/3 \rfloor = 15$. Divisible by 5: $\lfloor 46/5 \rfloor = 9$. Divisible by 15: $\lfloor 46/15 \rfloor = 3$. By Inclusion-Exclusion: $15 + 9 - 3 = 21$.
ANSWER 6: B

Problem 7:
Let $x$ be the 1st tank. 2nd: $x+1$, 3rd: $x+2$, 4th: $x+3$. Total: $x + (x+1) + (x+2) + (x+3) = 90 \implies 4x + 6 = 90 \implies 4x = 84 \implies x = 21$. 4th tank = $21+3 = 24$.
ANSWER 7: D

Problem 8:
Area $A = LW$. New $A' = (1.2L)(1.5W) = 1.8LW = 1.8A$. The increase is $1.8 - 1 = 0.8$, or 80%.
ANSWER 8: D

Problem 9:
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = (48)(290) = (16 \times 3)(29 \times 10) = 16 \times 3 \times 29 \times 2 \times 5 = 32 \times 3 \times 29 \times 5$. The largest power of 2 is 32.
ANSWER 9: C

Problem 10:
Let the square have vertices $(0,0), (10,0), (10,10), (0,10)$. Lemming starts at $(0,0)$. Diagonal vector is $(1,1)$, length $\sqrt{2} \approx 1.414$. $6.2/\sqrt{2} \approx 4.38$. Position after diagonal: $(4.38, 4.38)$. Turn 90 degrees: new vector $(4.38, -4.38)$ normalized is $(1/\sqrt{2}, -1/\sqrt{2})$. Move 2 meters: $(4.38 + 2/\sqrt{2}, 4.38 - 2/\sqrt{2}) \approx (4.38 + 1.41, 4.38 - 1.41) = (5.79, 2.97)$. Distances to sides: $x, 10-x, y, 10-y$. Average = $(x + 10-x + y + 10-y)/4 = 20/4 = 5$.
ANSWER 10: C

Problem 11:
Sum of $n$ terms in AP is $n \times (\text{average})$. $10000 / 25 = 400$. The middle term (13th term) is 400. The 25th term is $400 + 12 \times 2 = 424$.
ANSWER 11: E

Problem 12:
In 30 minutes of real time, the clock gains 5 minutes. Ratio: Clock runs at $35/30 = 7/6$ speed of real time. From 12:00 to 7:00 on the car clock is 7 hours (420 minutes). Let $t$ be real time: $t \times (7/6) = 420 \implies t = 420 \times 6 / 7 = 360$ minutes. 360 minutes = 6 hours. 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
The original number of license plates is \(5 \times 3 \times 4 = 60\). Adding two letters can increase one set by 2 or two different sets by 1 each. The resulting totals are:
- Both added to the first set: \(7 \times 3 \times 4 = 84\) (24 additional).
- Both added to the second set: \(5 \times 5 \times 4 = 100\) (40 additional).
- Both added to the third set: \(5 \times 3 \times 6 = 90\) (30 additional).
- One each to the first and second sets: \(6 \times 4 \times 4 = 96\) (36 additional).
- One each to the first and third sets: \(6 \times 3 \times 5 = 90\) (30 additional).
- One each to the second and third sets: \(5 \times 4 \times 5 = 100\) (40 additional).
No distribution yields more than 40 additional plates, so the maximum is 40 (choice D, not the tempting values 36 or 60).
ANSWER 1: D
Problem 2:
The trip covers \(57060 - 56200 = 860\) miles. The three fill-ups of 6, 12, and 20 gallons together replace all fuel consumed on the trip, for a total of 38 gallons. Thus, average mileage is \(860 / 38 \approx 22.6316\), which rounds to 22.6 (choice B).
ANSWER 2: B
Problem 3:
The mean of 90 implies a total sum of 450. The median of 91 forces the middle score (when ordered) to be 91. The unique mode of 94 requires 94 to appear strictly more often than any other value. The only configurations satisfying all three statistics produce different sums for the two lowest scores (171 when 94 appears twice, 168 when 94 appears three times). Because multiple values are possible, the sum is not uniquely determined (choice E).
ANSWER 3: E
Problem 4:
Label positions modulo 12, with 12 identified with 0. Alice's position after \(k\) turns is \(5k \pmod{12}\). Bob's position is \(-9k \equiv 3k \pmod{12}\). Setting them equal yields \(5k \equiv 3k \pmod{12}\), or \(2k \equiv 0 \pmod{12}\), so \(k\) is a multiple of 6. The smallest positive integer solution is therefore \(k = 6\) (choice A).
ANSWER 4: A
Problem 5:
For even \(n \ge 10\), \(n!!\) contains the factor 10 and therefore ends in the digit 0. The only terms whose units digits are nonzero are therefore \(2!!\), \(4!!\), \(6!!\), and \(8!!\), whose units digits are 2, 8, 8, and 4. Their sum has units digit 2; adding any number of zeros leaves the units digit unchanged. Hence the units digit of the whole sum is 2 (choice B).
ANSWER 5: B
Problem 6:
There are \(\lfloor 46/3 \rfloor = 15\) multiples of 3, \(\lfloor 46/5 \rfloor = 9\) multiples of 5, and \(\lfloor 46/15 \rfloor = 3\) multiples of 15. By inclusion-exclusion the number divisible by 3 or 5 is \(15 + 9 - 3 = 21\) (choice B).
ANSWER 6: B
Problem 7:
Let the first tank contain \(x\) guppies. Then the tanks contain \(x\), \(x+1\), \(x+3\), and \(x+6\) guppies. Their sum is \(4x + 10 = 90\), so \(x = 20\). The fourth tank therefore holds \(20 + 6 = 26\) guppies (choice E).
ANSWER 7: E
Problem 8:
A 20 % increase multiplies length by 1.2; a 50 % increase multiplies width by 1.5. The area is therefore multiplied by \(1.2 \times 1.5 = 1.8\), an 80 % increase (choice D).
ANSWER 8: D
Problem 9:
Factor the difference of fourth powers:
\[
13^4 - 11^4 = (13^2 + 11^2)(13 + 11)(13 - 11) = 290 \times 24 \times 2.
\]
Now \(290 = 2 \times 5 \times 29\) and \(24 = 2^3 \times 3\), so the prime factorization contains exactly five factors of 2. The highest power of 2 that divides the expression is therefore \(2^5 = 32\) (choice C).
ANSWER 9: C
Problem 10:
At any interior point \((x, y)\) the four distances to the sides of the square are \(x\), \(10 - x\), \(y\), and \(10 - y\). Their average is always
\[
\frac{x + (10 - x) + y + (10 - y)}{4} = 5
\]
(choice C). The explicit path taken by the lemming is irrelevant.
ANSWER 10: C
Problem 11:
The middle term of 25 consecutive even integers equals their average, \(10000 / 25 = 400\). The 13th term is therefore 400, and the 25th term (12 steps of +2 later) is \(400 + 24 = 424\) (choice E).
ANSWER 11: E
Problem 12:
In the 30 real minutes from noon to 12:30, the car clock advances 35 minutes and therefore gains 5 minutes. The constant gain rate is therefore \(5/30 = 1/6\) minute per real minute. When the car clock later reads 7:00 it has advanced 420 minutes, so the real time elapsed \(r\) satisfies
\[
r + \frac16 r = 420 \implies \frac76 r = 420 \implies r = 360
\]
minutes = 6 hours. The actual time is therefore 6:00 (choice B).
ANSWER 12: B

## Step 1: Problem 1 - Understand the initial conditions for license plates in Flatville.
The problem states that initially, bicycle license plates in Flatville are made of three letters. The first letter is chosen from 5 letters, the second from 3 letters, and the third from 4 letters. So, the total number of license plates initially is $5 \times 3 \times 4 = 60$.

## Step 2: Problem 1 - Determine the possible ways to add two new letters to the sets.
To maximize the number of additional license plates, we need to consider where adding two letters will result in the largest increase. The options are: adding 2 letters to the first set (making it $5+2 = 7$), adding 2 letters to the second set (making it $3+2 = 5$), adding 2 letters to the third set (making it $4+2 = 6$), or distributing the 2 letters across different sets.

## Step 3: Problem 1 - Calculate the total number of license plates for each scenario of adding 2 letters.
1. Adding 2 to the first set: $7 \times 3 \times 4 = 84$ plates, so $84 - 60 = 24$ additional plates.
2. Adding 2 to the second set: $5 \times 5 \times 4 = 100$ plates, so $100 - 60 = 40$ additional plates.
3. Adding 2 to the third set: $5 \times 3 \times 6 = 90$ plates, so $90 - 60 = 30$ additional plates.
4. Distributing 2 letters: We can add 1 letter to each of two sets. The possibilities are:
- First and second set: $(5+1) \times (3+1) \times 4 = 6 \times 4 \times 4 = 96$, so $96 - 60 = 36$ additional plates.
- First and third set: $(5+1) \times 3 \times (4+1) = 6 \times 3 \times 5 = 90$, so $90 - 60 = 30$ additional plates.
- Second and third set: $5 \times (3+1) \times (4+1) = 5 \times 4 \times 5 = 100$, so $100 - 60 = 40$ additional plates.

## Step 4: Problem 1 - Identify the maximum number of additional license plates.
From the calculations, the maximum number of additional plates is achieved by either adding 2 letters to the second set or by adding 1 letter to the second set and 1 to the third set, both resulting in 40 additional plates.

## Step 5: Problem 1 - Provide the final answer.
The largest possible number of additional license plates that can be made is 40.

ANSWER 1: D

## Step 6: Problem 2 - Understand the problem and the given conditions.
The driver starts with 56,200 miles, fills the tank with 6 gallons, then fills it again with 12 gallons at 56,560 miles, and finally fills it with 20 gallons at 57,060 miles.

## Step 7: Problem 2 - Calculate the total distance traveled and the total gasoline used.
Total distance = 57,060 - 56,200 = 860 miles. Total gasoline used = 6 + 12 + 20 = 38 gallons.

## Step 8: Problem 2 - Calculate the average miles-per-gallon for the entire trip.
Average miles-per-gallon = Total distance / Total gasoline used = 860 / 38.

## Step 9: Problem 2 - Perform the division to find the average miles-per-gallon.
860 / 38 = 22.631578947... To the nearest tenth, this is 22.6.

## Step 10: Problem 2 - Provide the final answer.
The average miles-per-gallon for the entire trip is 22.6.

ANSWER 2: B

## Step 11: Problem 3 - Understand the given conditions about the test scores.
The mean of 5 test scores is 90, the median is 91, and the mode is 94. This implies that the sum of all scores is $90 \times 5 = 450$.

## Step 12: Problem 3 - Use the given information to deduce the test scores.
Since the median is 91, and there are 5 scores, the third score must be 91. The mode is 94, meaning at least two scores are 94. So, we have at least two 94s. Let's assume the scores in ascending order are $a, b, 91, 94, 94$.

## Step 13: Problem 3 - Calculate the sum of the two lowest test scores.
The sum of all scores is 450, so $a + b + 91 + 94 + 94 = 450$. Simplifying, $a + b + 279 = 450$, hence $a + b = 450 - 279 = 171$.

## Step 14: Problem 3 - Provide the final answer.
The sum of the two lowest test scores is 171.

ANSWER 3: B

## Step 15: Problem 4 - Understand the movement of Alice and Bob.
Alice moves 5 points clockwise, and Bob moves 9 points counterclockwise. They start at point 12.

## Step 16: Problem 4 - Determine the condition for them to meet.
They will meet when they are at the same point. Since there are 12 points, we need to consider their relative positions after each turn.

## Step 17: Problem 4 - Analyze their movement.
After one turn, Alice is at $12 + 5 = 17 \equiv 5$ (mod 12), and Bob is at $12 - 9 = 3$. We need to find when they will be at the same point.

## Step 18: Problem 4 - Calculate the number of turns it takes for Alice and Bob to meet.
Let's track their positions: Alice's position after $n$ turns is $12 + 5n \mod 12$, and Bob's is $12 - 9n \mod 12$. We need $12 + 5n \equiv 12 - 9n \mod 12$, which simplifies to $14n \equiv 0 \mod 12$.

## Step 19: Problem 4 - Simplify the congruence.
$14n \equiv 0 \mod 12$ simplifies to $2n \equiv 0 \mod 12$, and further to $n \equiv 0 \mod 6$. The smallest positive $n$ satisfying this is $n = 6$.

## Step 20: Problem 4 - Provide the final answer.
It will take 6 turns for Alice and Bob to meet.

ANSWER 4: A

## Step 21: Problem 5 - Understand the double-factorial notation and its units digit for even n.
For $n \geq 10$, $n!!$ includes the product $2 \times 4 \times 6 \times 8 \times 10 = 3840$, which ends in 0. Thus, for $n \geq 10$, $n!!$ ends in 0.

## Step 22: Problem 5 - Calculate the units digit of the given sum.
We need to calculate $2!! + 4!! + 6!! + 8!! + 10!! + ... + 2022!!$. For $n \geq 10$, $n!!$ ends in 0. So, we only need to consider $2!! + 4!! + 6!! + 8!!$.

## Step 23: Problem 5 - Compute the sum of the double-factorials for n = 2, 4, 6, 8.
$2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$. The sum is $2 + 8 + 48 + 384 = 442$, which ends in 2.

## Step 24: Problem 5 - Provide the final answer.
The units digit of the sum is 2.

ANSWER 5: B

## Step 25: Problem 6 - Determine the number of whole numbers from 1 to 46 divisible by 3 or 5.
First, count the numbers divisible by 3: $\left\lfloor\frac{46}{3}\right\rfloor = 15$. Then, count the numbers divisible by 5: $\left\lfloor

**Problem 1:**
The original sets have sizes 5 (first letter), 3 (second), 4 (third). Total plates = \(5 \times 3 \times 4 = 60\).
We add 2 new letters, distributing them among the three sets to maximize the *increase* in the number of plates.  
Let the new sizes be \(a', b', c'\) with \(a'+b'+c' = (5+3+4)+2 = 14\). The increase is \(a'b'c' - 60\).
Check all ways to add two letters:
- Both to first: \(7 \times 3 \times 4 = 84 \rightarrow +24\)
- Both to second: \(5 \times 5 \times 4 = 100 \rightarrow +40\)
- Both to third: \(5 \times 3 \times 6 = 90 \rightarrow +30\)
- One to first, one to second: \(6 \times 4 \times 4 = 96 \rightarrow +36\)
- One to first, one to third: \(6 \times 3 \times 5 = 90 \rightarrow +30\)
- One to second, one to third: \(5 \times 4 \times 5 = 100 \rightarrow +40\)
The largest increase is 40.

ANSWER 1: D

**Problem 2:**
Trip mileage: start 56,200; mid 56,560; end 57,060.  
Total distance = \(57,060 - 56,200 = 860\) miles.
Gasoline added: 6 gal at start (to fill tank), 12 gal during trip (to fill tank again), 20 gal at end (to fill tank again).  
Since the tank was full at the start (after adding 6) and full at the end (after adding 20), the gasoline *used* during the trip is the sum of the intermediate fill‑ups = \(12 + 20 = 32\) gallons. (The initial 6 gal only replaces fuel already in the tank to make it full; it cancels out when comparing two full tanks.)
Average mpg = \(860 \div 32 = 26.875 \approx 26.9\).

ANSWER 2: D

**Problem 3:**
Five scores with mean 90 → sum = \(5 \times 90 = 450\).
Median = 91, so the third score (ordered) is 91.
Mode = 94, so 94 appears at least twice. Since scores are ordered \(a \le b \le c=91 \le d \le e\), the only way 94 appears at least twice is \(d = e = 94\) (cannot appear in positions 1 or 2 because they are \(\le 91\)).  
Sum: \(a + b + 91 + 94 + 94 = a + b + 279 = 450 \Rightarrow a + b = 171\).

ANSWER 3: B

**Problem 4:**
Number points 1–12 clockwise. Start at 12 (position 0 mod 12).  
Alice moves +5 clockwise each turn. Bob moves –9 counterclockwise ≡ +3 mod 12 each turn.  
They meet when \(5t \equiv 3t \pmod{12} \Rightarrow 2t \equiv 0 \pmod{12} \Rightarrow t \equiv 0 \pmod 6\).  
Smallest positive \(t\) is 6. (Check: after 6 turns both are at point 6.)  

ANSWER 4: A

**Problem 5:**
\(n!!\) for even \(n\) is product of even numbers from 2 to \(n\).  
For \(n \ge 10\), the product includes 10, so the units digit is 0.  
Sum units digit = units digit of \((2!! + 4!! + 6!! + 8!!)\).  
\(2!! = 2\), \(4!! = 8\), \(6!! = 48\) (units 8), \(8!! = 384\) (units 4).  
Sum = \(2 + 8 + 8 + 4 = 22 \rightarrow\) units digit 2.

ANSWER 5: B

**Problem 6:**
Numbers 1 through 46 divisible by 3 or 5.  
Count = \(\lfloor 46/3\rfloor + \lfloor 46/5\rfloor - \lfloor 46/15\rfloor = 15 + 9 - 3 = 21\).

ANSWER 6: B

**Problem 7:**
Let first tank have \(x\) guppies.  
Second: \(x+1\)  
Third: \((x+1)+2 = x+3\)  
Fourth: \((x+3)+3 = x+6\)  
Total: \(x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90 \Rightarrow 4x = 80 \Rightarrow x = 20\).  
Fourth tank = \(20 + 6 = 26\).

ANSWER 7: E

**Problem 8:**
Original area = \(L \times W\).  
New length = \(1.20L\), new width = \(1.50W\).  
New area = \(1.20 \times 1.50 \, LW = 1.80 \, LW\).  
Increase = \(80\%\).

ANSWER 8: D

**Problem 9:**
\(13^4 - 11^4 = (13^2-11^2)(13^2+11^2) = (13-11)(13+11)(169+121)\)  
\(= 2 \times 24 \times 290\).  
\(24 = 2^3 \times 3\), so \(2 \times 24 = 2^4 \times 3 = 48\).  
\(290 = 2 \times 145 = 2 \times 5 \times 29\).  
Product = \(2^4 \times 3 \times 2 \times 5 \times 29 = 2^5 \times (3 \times 5 \times 29) = 32 \times 435\).  
Largest power of 2 divisor is \(32\).

ANSWER 9: C

**Problem 10:**
Place square corners at \((0,0), (10,0), (10,10), (0,10)\). Start at \((0,0)\).  
Movement: 6.2 m along diagonal toward \((10,10)\). Diagonal unit vector = \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\); after this step, coordinates = \((\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}) \approx (4.384, 4.384)\).  
Right turn → new direction is \((\frac{1}{\sqrt2}, -\frac{1}{\sqrt2})\) (or equivalent). Run 2 m: displacement = \((\sqrt2, -\sqrt2) \approx (1.414, -1.414)\).  
Final coordinates = \((4.384+1.414,\; 4.384-1.414) = (5.798, 2.97)\) — inside the square.  
Shortest distances to the four sides: left \(x\), right \(10-x\), bottom \(y\), top \(10-y\).  
Sum = \(x + (10-x) + y + (10-y) = 20\).  
Average = \(20/4 = 5\).

ANSWER 10: C

**Problem 11:**
Let the 25 consecutive even integers be \(a, a+2, \dots, a+48\).  
Sum = \(\frac{25}{2}(a + a+48) = 25(a+24) = 10,000 \Rightarrow a+24 = 400 \Rightarrow a = 376\).  
Largest = \(a + 48 = 424\).

ANSWER 11: E

**Problem 12:**
In 30 real minutes, the car clock advances 35 minutes → car clock runs at \(\frac{35}{30} = \frac{7}{6}\) times real speed.  
When car clock reads 7:00 (420 minutes after 12:00), real time elapsed \(T\) satisfies \(\frac{7}{6}T = 420 \Rightarrow T = 420 \times \frac{6}{7} = 360\) minutes = 6 hours.  
Actual time = 12:00 + 6:00 = 6:00.

ANSWER 12: B

# Problem 1

**Restating:** We have three sets of letters with sizes 5, 3, and 4. The original number of plates is $5 \times 3 \times 4 = 60$. We add 2 letters total (distributed among the sets however we like) and want to maximize the *additional* plates.

**Setting up:** Let the new set sizes be $a, b, c$ where $a + b + c = 5 + 3 + 4 + 2 = 14$. We want to maximize $a \times b \times c$.

**Testing all distributions of 2 extra letters:**
- Both to Set 1: $7 \times 3 \times 4 = 84 \implies$ additional = 24
- Both to Set 2: $5 \times 5 \times 4 = 100 \implies$ additional = 40
- Both to Set 3: $5 \times 3 \times 6 = 90 \implies$ additional = 30
- One to Set 1, one to Set 2: $6 \times 4 \times 4 = 96 \implies$ additional = 36
- One to Set 1, one to Set 3: $6 \times 3 \times 5 = 90 \implies$ additional = 30
- One to Set 2, one to Set 3: $5 \times 4 \times 5 = 100 \implies$ additional = 40

**Ruling out wrong choices:** Adding both letters to the smallest set (Set 2, size 3) or splitting one each to Sets 2 and 3 gives the maximum product of 100, yielding $100 - 60 = 40$ additional plates. Choices A, B, C, and E correspond to suboptimal distributions or arithmetic errors.

ANSWER 1: D

---

# Problem 2

**Restating:** We need the average miles-per-gallon for the entire trip.

**Setting up:**
- Trip starts at odometer 56,200 (tank filled with 6 gal — this just tops off the tank before driving).
- At 56,560, the driver refills 12 gal — this is the gas consumed from 56,200 to 56,560.
- At 57,060, the driver refills 20 gal — this is the gas consumed from 56,560 to 57,060.

**Computing:**
- Total distance = $57{,}060 - 56{,}200 = 860$ miles
- Total gas consumed = $12 + 20 = 32$ gallons (the initial 6 gallons just filled the tank before the trip began; the refills measure actual consumption)
- Average MPG = $\frac{860}{32} = 26.875$

Rounding to the nearest tenth: $26.9$.

**Ruling out wrong choices:** Choice A (22.5) or B (22.6) might come from incorrectly including the 6 gallons ($860/38 \approx 22.6$). Choice E (27.5) might come from using only part of the trip. The correct total gas used is 32 gallons.

ANSWER 2: D

---

# Problem 3

**Restating:** Five scores $a \le b \le c \le d \le e$ have mean 90, median 91, and mode 94. Find $a + b$.

**Setting up:**
- Mean = 90 $\implies a + b + c + d + e = 450$
- Median = 91 $\implies c = 91$
- Mode = 94 $\implies$ 94 appears more often than any other score

**Reasoning:** Since $c = 91$ and scores are ordered, 94 can only appear in positions $d$ and $e$. For 94 to be the mode, it must appear at least twice, so $d = e = 94$. (It can't appear three times since $c = 91 \neq 94$.)

**Computing:**
$$a + b + 91 + 94 + 94 = 450$$
$$a + b = 450 - 279 = 171$$

**Ruling out wrong choices:** Even if $a = b$ (giving them a frequency of 2, tied with 94), the sum $a + b$ is still forced to be 171. The answer is fully determined, ruling out choice E.

ANSWER 3: B

---

# Problem 4

**Restating:** On a 12-point circle, Alice moves +5 (clockwise) and Bob moves −9 (counterclockwise) each turn. Both start at 12. When do they land on the same point?

**Setting up:** After $t$ turns:
- Alice's position: $5t \pmod{12}$
- Bob's position: $-9t \equiv 3t \pmod{12}$

They meet when $5t \equiv 3t \pmod{12}$, i.e., $2t \equiv 0 \pmod{12}$.

**Computing:** $2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}$. The smallest positive $t$ is $6$.

**Verification:** After 6 turns, Alice is at $5 \times 6 = 30 \equiv 6 \pmod{12}$ and Bob is at $-9 \times 6 = -54 \equiv 6 \pmod{12}$. Both are at point 6. ✓

**Ruling out wrong choices:** Choices B, C, D, E are larger multiples or incorrect modular arithmetic results.

ANSWER 4: A

---

# Problem 5

**Restating:** Find the units digit of $2!! + 4!! + 6!! + \cdots + 2022!!$.

**Computing the first few double factorials:**
- $2!! = 2$ → units digit **2**
- $4!! = 2 \times 4 = 8$ → units digit **8**
- $6!! = 2 \times 4 \times 6 = 48$ → units digit **8**
- $8!! = 48 \times 8 = 384$ → units digit **4**
- $10!! = 384 \times 10 = 3840$ → units digit **0**

**Key insight:** For all even $n \ge 10$, $n!!$ includes both 2 and 5 as factors (since $2$ and $10$ are both $\le n$), so $n!!$ is a multiple of 10, giving units digit **0**.

**Summing the relevant units digits:**
$$2 + 8 + 8 + 4 + 0 + 0 + \cdots = 22$$

The units digit of the total sum is **2**.

ANSWER 5: B

---

# Problem 6

**Restating:** Count integers from 1 to 46 divisible by 3 or 5 (or both).

**Using inclusion-exclusion:**
- Divisible by 3: $\lfloor 46/3 \rfloor = 15$
- Divisible by 5: $\lfloor 46/5 \rfloor = 9$
- Divisible by both (i.e., by 15): $\lfloor 46/15 \rfloor = 3$

$$\text{Total} = 15 + 9 - 3 = 21$$

**Ruling out wrong choices:** Choice A (18) forgets to add back the overlap or miscounts. Choice D (25) double-counts the multiples of 15.

ANSWER 6: B

---

# Problem 7

**Restating:** Four tanks hold 90 guppies total, with successive differences of +1, +2, +3.

**Setting up:** Let the 1st tank have $x$ guppies.
- 2nd tank: $x + 1$
- 3rd tank: $x + 1 + 2 = x + 3$
- 4th tank: $x + 3 + 3 = x + 6$

**Computing:**
$$x + (x+1) + (x+3) + (x+6) = 90$$
$$4x + 10 = 90 \implies 4x = 80 \implies x = 20$$

4th tank: $x + 6 = 20 + 6 = 26$.

**Verification:** $20 + 21 + 23 + 26 = 90$ ✓

ANSWER 7: E

---

# Problem 8

**Restating:** Length increases by 20%, width by 50%. What is the percent increase in area?

**Setting up:**
- Original area: $A = L \times W$
- New area: $A' = (1.2L)(1.5W) = 1.8LW$

**Computing:**
$$\text{Increase} = \frac{1.8LW - LW}{LW} = 0.8 = 80\%$$

**Ruling out wrong choices:** Choice B (30%) is the trap of simply adding 20% + 50% and not accounting for the cross term. Choice C (70%) might come from $20\% + 50\% = 70\%$ without the interaction. The correct calculation gives $1.2 \times 1.5 = 1.8$, an 80% increase.

ANSWER 8: D

---

# Problem 9

**Restating:** Find the largest power of 2 dividing $13^4 - 11^4$.

**Factoring using difference of squares:**
$$13^4 - 11^4 = (

**Problem 1:**
A Flatville license plate has three letters. The first letter is chosen from 5 options, the second from 3 options, and the third from 4 options, so the current number of plates is \(5 \times 3 \times 4 = 60\). We are adding two new letters to the sets, either both to one set or split between two sets. We want the largest possible increase in the number of plates.

- Add both to the first set: \((5+2) \times 3 \times 4 = 84\), additional plates \(= 24\).
- Add both to the second set: \(5 \times (3+2) \times 4 = 100\), additional plates \(= 40\).
- Add both to the third set: \(5 \times 3 \times (4+2) = 90\), additional plates \(= 30\).
- Add one to the first and one to the second: \(6 \times 4 \times 4 = 96\), additional plates \(= 36\).
- Add one to the first and one to the third: \(6 \times 3 \times 5 = 90\), additional plates \(= 30\).
- Add one to the second and one to the third: \(5 \times 4 \times 5 = 100\), additional plates \(= 40\).

The maximum number of additional plates is \(40\). The values \(24\), \(30\), and \(36\) are too small, and \(60\) is not attainable.

ANSWER 1: D

---

**Problem 2:**
The car’s total distance traveled is the difference between the final and initial odometer readings:
\[
57{,}060 - 56{,}200 = 860 \text{ miles}.
\]

To find the gasoline used **during the trip**, note that the amount added at a refueling equals the amount burned since the previous fill-up (assuming the tank is filled to capacity each time). The \(12\) gallons added during the trip replaced the fuel burned in the first segment, and the \(20\) gallons added at the end replaced the fuel burned in the second segment. The initial \(6\) gallons was added before the measured trip began and does not count as fuel consumed during the driving.

\[
\text{Total fuel used} = 12 + 20 = 32 \text{ gallons}.
\]

\[
\text{Average mpg} = \frac{860}{32} = 26.875 \approx 26.9.
\]

The choice \(22.6\) results from mistakenly including the initial \(6\) gallons.

ANSWER 2: D

---

**Problem 3:**
Let the five test scores in non-decreasing order be \(a \le b \le c \le d \le e\).

- Mean \(90\): \(a+b+c+d+e = 5 \times 90 = 450\).
- Median \(91\): the middle score \(c = 91\).
- Mode \(94\): \(94\) appears most frequently. Since \(c=91\), the only way \(94\) can be the unique mode is if \(d = e = 94\) and no other score repeats. Thus \(a < b < 91\).

Substituting:
\[
a + b + 91 + 94 + 94 = 450
\]
\[
a + b + 279 = 450
\]
\[
a + b = 171.
\]

The sum of the two lowest scores is \(171\). The information is sufficient, so “not determined” is incorrect.

ANSWER 3: B

---

**Problem 4:**
Number the points \(0,1,2,\dots,11\) clockwise, with \(12\) as \(0\).

- Alice moves \(5\) points clockwise each turn, so after \(t\) turns she is at position \(5t \pmod{12}\).
- Bob moves \(9\) points counterclockwise. Since \(9\) counterclockwise is equivalent to \(3\) clockwise modulo \(12\), after \(t\) turns he is at position \(3t \pmod{12}\).

They meet when
\[
5t \equiv 3t \pmod{12} \implies 2t \equiv 0 \pmod{12} \implies t \equiv 0 \pmod{6}.
\]

The smallest positive integer \(t\) is \(6\). (Check: Alice is at \(5 \times 6 = 30 \equiv 6\); Bob is at \(3 \times 6 = 18 \equiv 6\).)

ANSWER 4: A

---

**Problem 5:**
We need the units digit of \(2!! + 4!! + 6!! + \dots + 2022!!\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit \(2\)
- \(4!! = 2 \times 4 = 8\) → units digit \(8\)
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit \(8\)
- \(8!! = 48 \times 8 = 384\) → units digit \(4\)
- \(10!! = 384 \times 10 = 3840\) → units digit \(0\)

For every even \(n \ge 10\), the product \(n!!\) contains a factor of \(10\), so its units digit is \(0\).

Adding the relevant units digits:
\[
2 + 8 + 8 + 4 = 22,
\]
which has units digit \(2\).

ANSWER 5: B

---

**Problem 6:**
Count numbers from \(1\) to \(46\) divisible by \(3\) or \(5\).

- Divisible by \(3\): \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Divisible by \(5\): \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Divisible by both (i.e., by \(15\)): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

**Problem 7:**
Let the number of guppies in the 1st tank be \(x\).

- 2nd tank: \(x+1\)
- 3rd tank: \((x+1)+2 = x+3\)
- 4th tank: \((x+3)+3 = x+6\)

Total:
\[
x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90
\]
\[
4x = 80 \implies x = 20.
\]

The 4th tank has \(20 + 6 = 26\) guppies.

ANSWER 7: E

---

**Problem 8:**
Let the original length be \(L\) and width be \(W\). Original area \(= LW\).

New dimensions:
- Length \(= 1.2L\)
- Width \(= 1.5W\)

New area:
\[
(1.2L)(1.5W) = 1.8LW.
\]

This is \(180\%\) of the original area, so the area has increased by \(80\%\).

ANSWER 8: D

---

**Problem 9:**
Factor \(13^4 - 11^4\) using difference of squares twice:

\[
13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 - 11^2)(13^2 + 11^2).
\]

\[
13^2 - 11^2 = (13-11)(13+11) = 2 \times 24 = 48 = 2^4 \cdot 3.
\]

\[
13^2 + 11^2 = 169 + 121 = 290 = 2 \cdot 145.
\]

Multiplying:
\[
13^4 - 11^4 = (2^4 \cdot 3)(2 \cdot 145) = 2^5 \cdot 3 \cdot 145.
\]

The highest power of \(2\) dividing this is \(2^5 = 32\).

ANSWER 9: C

---

**Problem 10:**
Place the square with corners at \((0,0)\), \((10,0)\), \((10,10)\), \((0,10)\). The lemming starts at \((0,0)\) and runs \(6.2\) m toward \((10,10)\).

The diagonal direction is \((1,1)\). After \(6.2\) m, the coordinates are
\[
\left(\frac{6.2}{\sqrt2}, \frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\, 3.1\sqrt2).
\]

A \(90^\circ\) right turn from direction \((1,1)\) gives direction \((1,-1)\). Running \(2\) m in this direction adds \((\sqrt2,\, -\sqrt2)\). The final position is
\[
(4.1\sqrt2,\, 2.1\sqrt2).
\]

The four shortest distances to the sides are:
- to \(x=0\): \(x\)
- to \(x=10\): \(10-x\)
- to \(y=0\): \(y\)
- to \(y=10\): \(10-y\)

Their sum is
\[
x + (10-x) + y + (10-y) = 20.
\]

The average is
\[
\frac{20}{4} = 5.
\]

ANSWER 10: C

---

**Problem 11:**
For \(25\) co

Problem 1:
We need to find the maximum number of additional license plates when adding 2 new letters to the three sets.

**Initial Setup:**
- Set 1: {C, H, L, P, R} → 5 letters
- Set 2: {A, I, O} → 3 letters
- Set 3: {D, M, N, T} → 4 letters
- Initial plates = 5 × 3 × 4 = 60

**Adding both letters to one set:**
- Both to Set 1: 7 × 3 × 4 = 84, additional = 84 - 60 = 24
- Both to Set 2: 5 × 5 × 4 = 100, additional = 100 - 60 = 40
- Both to Set 3: 5 × 3 × 6 = 90, additional = 90 - 60 = 30

**Adding one letter to each of two sets:**
- Sets 1 & 2: 6 × 4 × 4 = 96, additional = 96 - 60 = 36
- Sets 1 & 3: 6 × 3 × 5 = 90, additional = 90 - 60 = 30
- Sets 2 & 3: 5 × 4 × 5 = 100, additional = 100 - 60 = 40

The largest additional number is 40, achieved by adding both letters to Set 2, or one to Set 2 and one to Set 3.

ANSWER 1: D

---

Problem 2:
We need to find the average miles-per-gallon for the entire trip.

**Total distance traveled:**
57,060 - 56,200 = 860 miles

**Total gasoline used:**
The driver filled the tank at the start (6 gal), during the trip (12 gal), and at the end (20 gal). All gasoline purchased was consumed during the trip.
Total = 6 + 12 + 20 = 38 gallons

**Average mpg:**
860 ÷ 38 = 22.6315... ≈ 22.6

Note: The intermediate odometer reading (56,560) is a distractor.

ANSWER 2: B

---

Problem 3:
We have five test scores with mean 90, median 91, and mode 94.

**From the mean:** Sum of all 5 scores = 5 × 90 = 450

**From the median:** When sorted, the 3rd score = 91

**From the mode:** 94 appears most frequently. Since the median is 91, at least two scores must be ≥ 91. For 94 to be the mode, it must appear at least twice.

Let the sorted scores be: a ≤ b ≤ 91 ≤ d ≤ e

For 94 to be the mode, we need at least two scores equal to 94. Since 91 is the median, the two largest could be 94 and 94:
a + b + 91 + 94 + 94 = 450
a + b = 171

Could the mode be achieved differently? If three scores were 94, then the median would need to be 94, contradicting median = 91. So exactly two scores are 94.

Could d > 94? Then we'd need another 94 somewhere, but if a or b = 94, then with d > 94 and e ≥ d, we'd have at most two 94s (same as before), giving a + b = 171 still.

The sum of the two lowest scores is 171.

ANSWER 3: B

---

Problem 4:
Alice and Bob start at point 12. Alice moves 5 points clockwise each turn; Bob moves 9 points counterclockwise each turn.

**Positions after n turns:**
- Alice: 12 + 5n (mod 12)
- Bob: 12 - 9n (mod 12)

They meet when: 12 + 5n ≡ 12 - 9n (mod 12)
This simplifies to: 14n ≡ 0 (mod 12)
Which means: 7n ≡ 0 (mod 6)

Since gcd(7, 6) = 1, we need n ≡ 0 (mod 6).

The smallest positive n is **6**.

**Verification:**
- After 6 turns, Alice is at 12 + 30 = 42 ≡ 6 (mod 12), so point 6
- After 6 turns, Bob is at 12 - 54 = -42 ≡ -42 + 48 = 6 (mod 12), so point 6 ✓

ANSWER 4: A

---

Problem 5:
We need the units digit of 2!! + 4!! + 6!! + ... + 2022!!

**Computing double factorials:**
- 2!! = 2
- 4!! = 2 × 4 = 8
- 6!! = 2 × 4 × 6 = 48
- 8!! = 2 × 4 × 6 × 8 = 384
- 10!! = 2 × 4 × 6 × 8 × 10 = 3840
- For n ≥ 10, n!! contains factors 2, 4, 6, 8, and 10, so it's divisible by 10, meaning units digit = 0

**Units digit sum:**
- 2!! → 2
- 4!! → 8
- 6!! → 8
- 8!! → 4
- 10!! through 2022!! → all contribute 0

Total units digit = 2 + 8 + 8 + 4 = 22, so units digit = **2**

ANSWER 5: B

---

Problem 6:
Count whole numbers from 1 to 46 divisible by 3 or 5.

**Divisible by 3:** ⌊46/3⌋ = 15 numbers
**Divisible by 5:** ⌊46/5⌋ = 9 numbers
**Divisible by both (lcm = 15):** ⌊46/15⌋ = 3 numbers

By inclusion-exclusion: 15 + 9 - 3 = **21**

ANSWER 6: B

---

Problem 7:
Let the number of guppies in the 1st tank be x.

- 1st tank: x
- 2nd tank: x + 1
- 3rd tank: x + 3
- 4th tank: x + 6

Total: x + (x + 1) + (x + 3) + (x + 6) = 4x + 10 = 90

4x = 80, so x = 20

4th tank: x + 6 = 20 + 6 = **26**

ANSWER 7: E

---

Problem 8:
Let original length = L and original width = W.
Original area = L × W

New length = 1.20L, new width = 1.50W
New area = 1.20L × 1.50W = 1.80LW

Increase = 1.80Lw - LW = 0.80LW, which is an **80%** increase.

ANSWER 8: D

---

Problem 9:
Find the largest power of 2 dividing 13⁴ - 11⁴.

**Factoring:**
13⁴ - 11⁴ = (13² - 11²)(13² + 11²) = (13 - 11)(13 + 11)(169 + 121)

= (2)(24)(290)

**Counting factors of 2:**
- 2 contributes 2¹
- 24 = 2³ × 3 contributes 2³
- 290 = 2 × 145 contributes 2¹

Total power of 2 = 1 + 3 + 1 = 5

So the largest power of 2 dividing 13⁴ - 11⁴ is 2⁵ = **32**.

ANSWER 9: C

---

Problem 10:
Place the square with corners at (0,0), (10,0), (10,10), (0,10). The lemming starts at (0,0) and runs along the diagonal toward (10,10).

**Diagonal direction:** The unit vector is (1/√2, 1/√2).

After running 6.2 meters along the diagonal:
Position = (6.2/√2, 6.2/√2) = (4.384, 4.384)

**Making a 90° right turn:** A right turn from the direction (1/√2, 1/√2) gives direction (1/√2, -1/√2).

Running 2 more meters:
Position = (4.384 + 2/√2, 4.384 - 2/√2) = (4.384 + 1.414, 4.384 - 1.414) = (5.798, 2.970)

**Distances to each side:**
- To bottom (y = 0): 2.970
- To top (y = 10): 10 - 2.970 = 7.030
- To left (x = 0): 5.798
- To right (x = 10): 10 - 5.798 = 4.202

**Average:** (2.970 + 7.030 + 5.798 + 4.202) / 4 = 20 / 4 = **5**

This makes sense: for any point inside a square with side s, the average distance to the four sides is always s/2 = 5.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be centered around their mean.

The mean = 10000/25 = 400, so the middle term (13th) is 400.

The integers are: 400 - 24, 400 - 22, ..., 400, ..., 400 + 24

The largest = 400 + 24 = **424**

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate relative to real time.

**Finding the rate:**
- Real time elapsed: 30 minutes (12:00 to 12:30)
- Car clock elapsed: 35 minutes (12:00 to 12:35)
- Rate: car clock runs at 35/30 = 7/6 times real speed

**When car clock shows 7:00:**
- Car clock elapsed from noon: 7 hours = 420 minutes
- Real time elapsed: 420

Problem 1:
The original license‑plate system has three independent choices:
- First letter: 5 possibilities,
- Second letter: 3 possibilities,
- Third letter: 4 possibilities.

Total plates originally = \(5 \times 3 \times 4 = 60\).

Two new letters are added.  They can be placed either both in one set or one in each of two sets.  We examine each distribution:

| Distribution | New sizes | New total plates | Increase |
|---|---|---|---|
| Both to first set | \(7,3,4\) | \(7\cdot3\cdot4 = 84\) | \(24\) |
| Both to second set | \(5,5,4\) | \(5\cdot5\cdot4 = 100\) | \(40\) |
| Both to third set | \(5,3,6\) | \(5\cdot3\cdot6 = 90\) | \(30\) |
| One to first, one to second | \(6,4,4\) | \(6\cdot4\cdot4 = 96\) | \(36\) |
| One to first, one to third | \(6,3,5\) | \(6\cdot3\cdot5 = 90\) | \(30\) |
| One to second, one to third | \(5,4,5\) | \(5\cdot4\cdot5 = 100\) | \(40\) |

The greatest increase is \(40\) plates (cases where both letters go to the second set, or one each to the second and third sets).

ANSWER 1: D

---

Problem 2:
Odometer readings: start \(56{,}200\) mi, end \(57{,}060\) mi.  
Total miles traveled = \(57{,}060 - 56{,}200 = 860\) mi.

Fuel added:
- At the start: 6 gal
- During the trip: 12 gal
- At the end: 20 gal  

Total gallons used = \(6 + 12 + 20 = 38\) gal.

Average miles‑per‑gallon = \(\dfrac{860}{38} \approx 22.6316\).  
Rounded to the nearest tenth: \(22.6\).

ANSWER 2: B

---

Problem 3:
Let the five test scores be \(a \le b \le c \le d \le e\).

- Mean = 90 ⇒ total = \(5 \times 90 = 450\).
- Median \(c = 91\).
- Mode = 94 ⇒ at least two scores are 94. Since the median is 91, the two 94’s must be the fourth and fifth scores: \(d = e = 94\).

Thus  
\(a + b + 91 + 94 + 94 = 450\)  
\(\Rightarrow a + b = 450 - 279 = 171\).

The sum of the two lowest scores is \(171\).

ANSWER 3: B

---

Problem 4:
Points are numbered modulo 12.  Starting at point 12:

- Alice moves \(+5\) points each turn.
- Bob moves \(-9\) points each turn, which is equivalent to \(+3\) points (since \(-9 \equiv 3 \pmod{12}\)).

After \(t\) turns:
\[
\text{Alice: } 12 + 5t \pmod{12},\qquad
\text{Bob: } 12 - 9t \pmod{12}.
\]

They meet when \(5t \equiv -9t \pmod{12}\), i.e. \(14t \equiv 0 \pmod{12}\).  
\(14 \equiv 2 \pmod{12}\), so \(2t \equiv 0 \pmod{12}\) ⇒ \(t \equiv 0 \pmod{6}\).

The smallest positive number of turns is \(t = 6\).

ANSWER 4: A

---

Problem 5:
For even \(n\), \(n!! = 2 \times 4 \times 6 \times \dots \times n\).

Compute the units digit of each term:
- \(2!! = 2\) → units digit 2
- \(4!! = 2 \times 4 = 8\) → units digit 8
- \(6!! = 2 \times 4 \times 6 = 48\) → units digit 8
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) → units digit 4
- For \(n \ge 10\), the product contains the factor 10, so the units digit is 0.

Thus the sum of units digits of all terms is  
\(2 + 8 + 8 + 4 = 22\), whose units digit is 2.

ANSWER 5: B

---

Problem 6:
Count integers from 1 to 46 divisible by 3 or 5 (or both).

- Multiples of 3: \(\left\lfloor \frac{46}{3} \right\rfloor = 15\).
- Multiples of 5: \(\left\lfloor \frac{46}{5} \right\rfloor = 9\).
- Multiples of both (i.e., 15): \(\left\lfloor \frac{46}{15} \right\rfloor = 3\).

By inclusion–exclusion:
\[
15 + 9 - 3 = 21.
\]

ANSWER 6: B

---

Problem 7:
Let the numbers of guppies in the four tanks be \(a, b, c, d\).

\[
\begin{cases}
b = a + 1,\\
c = b + 2 = a + 3,\\
d = c + 3 = a + 6,\\
a + b + c + d = 90.
\end{cases}
\]

Substituting:
\[
a + (a+1) + (a+3) + (a+6) = 90 \\
4a + 10 = 90 \\
4a = 80 \\
a = 20.
\]

Then \(d = a + 6 = 26\).

ANSWER 7: E

---

Problem 8:
Let original length = \(L\) and width = \(W\).

New length = \(L \times 1.20\), new width = \(W \times 1.50\).

New area = \(1.20 \times 1.50 \times L W = 1.80 \, L W\).

The area is increased by \(1.80 - 1 = 0.80\), i.e., \(80\%\).

ANSWER 8: D

---

Problem 9:
\[
13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (13-11)(13+11)(13^2+11^2).
\]

Compute each factor:
- \(13-11 = 2\),
- \(13+11 = 24\),
- \(13^2 + 11^2 = 169 + 121 = 290\).

Thus \(13^4 - 11^4 = 2 \times 24 \times 290 = 13920\).

Factor out powers of 2:
\[
13920 = 2^5 \times 435,
\]
and \(435\) is odd. Hence the largest power of 2 dividing the number is \(2^5 = 32\).

ANSWER 9: C

---

Problem 10:
Place the square with vertices \((0,0)\) to \((10,10)\).  
The lemming starts at \((0,0)\), runs \(6.2\) m along the diagonal toward \((10,10)\).  
The diagonal length is \(10\sqrt2\) m, so after this leg the coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right) = (3.1\sqrt2,\;3.1\sqrt2).
\]

A right turn (clockwise) changes direction to \((1,-1)\). Moving \(2\) m gives a displacement of \((\sqrt2,\,-\sqrt2)\).  
Final coordinates:
\[
x = 3.1\sqrt2 + \sqrt2 = 4.1\sqrt2,\qquad
y = 3.1\sqrt2 - \sqrt2 = 2.1\sqrt2.
\]

Distances to the four sides:
- Left side (\(x=0\)): \(x = 4.1\sqrt2\),
- Right side (\(x=10\)): \(10 - x = 10 - 4.1\sqrt2\),
- Bottom side (\(y=0\)): \(y = 2.1\sqrt2\),
- Top side (\(y=10\)): \(10 - y = 10 - 2.1\sqrt2\).

The sum of the four distances is
\[
x + (10-x) + y + (10-y) = 20,
\]
so the average is \(20/4 = 5\) meters.

ANSWER 10: C

---

Problem 11:
Let the 25 consecutive even integers be
\[
a,\; a+2,\; a+4,\; \dots,\; a+48.
\]

Their sum is
\[
\frac{25\,(a + (a+48))}{2} = 25\,(a+24) = 10{,}000.
\]

Thus \(a+24 = 10{,}000/25 = 400\) ⇒ \(a = 376\).  
The largest integer is \(a+48 = 376+48 = 424\).

ANSWER 11: E

---

Problem 12:
The car clock gains time at a constant rate. From the first two observations:

- Real time elapsed: 30 min (watch from 12:00 to 12:30).
- Car clock advanced: 35 min (from 12:00 to 12:35).

Hence the car clock runs at the rate \(\frac{35}{30} = \frac{7}{6}\) of real time.

When the car clock reads 7:00, that corresponds to \(7 \times 60 = 420\) minutes of car‑clock time.  
Let \(t\) be the real time in minutes after noon. Then
\[
420 = \frac{7}{6}\,t \quad\Longrightarrow\quad t = 420 \times \frac{6}{7} = 360\text{ minutes}.
\]

\(360\) minutes after noon is \(6\) hours later, i

Problem 1:  
Bicycle license plates in Flatville each contain three letters. The first letter is chosen from $\{C, H, L, P, R\}$ (5 options), the second from $\{A, I, O\}$ (3 options), and the third from $\{D, M, N, T\}$ (4 options). The total number of possible plates is $5 \times 3 \times 4 = 60$.  
Two new letters are added. The goal is to maximize the additional plates. Adding both letters to the second set (which has the fewest options, 3) increases its size to 5, giving $5 \times 5 \times 4 = 100$ plates, an increase of $100 - 60 = 40$. Alternatively, adding one letter to the second set (to 4) and one to the third set (to 5) gives $5 \times 4 \times 5 = 100$, also an increase of 40. Other distributions yield smaller increases.  
ANSWER 1: D  

Problem 2:  
The total distance traveled is $57,060 - 56,200 = 860$ miles. The total gasoline used is $6 + 12 + 20 = 38$ gallons (the initial 6 gallons are part of the trip, as the tank was filled before starting). The average miles per gallon is $860 / 38 \approx 22.6316$, which rounds to 22.6.  
ANSWER 2: C (Note: Re-evaluating the gas used—only the gas consumed during the trip matters. The first fill-up (6 gallons) is before the trip starts, so only the 12 gallons and 20 gallons are used for the distances between odometers. However, the trip includes the entire distance from 56,200 to 57,060, and the gas added during and after is all used. The total gas used for the trip is 12 + 20 = 32 gallons (the initial 6 gallons is not consumed during the trip). The distance is 57,060 - 56,200 = 860 miles. Thus, 860 / 32 = 26.875 ≈ 26.9. The correct calculation uses only the gas added during and after the trip for the distance covered.)  
Correction: Total gas used for the trip is 12 gallons (to go from 56,200 to 56,560, 360 miles) and 20 gallons (to go from 56,560 to 57,060, 500 miles), so total gas = 12 + 20 = 32 gallons, total distance = 860 miles, average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (original) is incorrect; correct is 26.9.  
Revised ANSWER 2: C (no, 26.9 is D).  
Final: Total distance = 57,060 - 56,200 = 860 miles. Gas used: first leg (56,200 to 56,560: 360 miles) used 12 gallons, second leg (56,560 to 57,060: 500 miles) used 20 gallons, total gas = 32 gallons. Average = 860 / 32 = 26.875 ≈ 26.9.  
ANSWER 2: C (no) D  

Problem 3:  
The mean of five test scores is 90, so the total sum is $5 \times 90 = 450$. The median is 91, so the third score is 91. The mode is 94, so 94 appears at least twice. To maximize the sum of the two lowest scores, minimize the higher scores. Let the scores be $a, b, 91, 94, 94$ (since mode is 94, and to minimize the higher scores, use two 94s). Then $a + b + 91 + 94 + 94 = 450$, so $a + b = 450 - 279 = 171$.  
ANSWER 3: B  

Problem 4:  
Alice starts at point 12 and moves 5 points clockwise each turn. Bob starts at point 12 and moves 9 points counterclockwise each turn. The circle has 12 points, so positions are modulo 12. Alice's position after $n$ turns: $(12 + 5n) \mod 12$. Bob's position: $(12 - 9n) \mod 12$. Set equal: $(5n) \mod 12 = (-9n) \mod 12$, or $5n \equiv 3n \mod 12$ (since -9 mod 12 is 3), so $2n \equiv 0 \mod 12$, so $n$ is a multiple of 6. The smallest positive $n$ is 6.  
ANSWER 4: A  

Problem 5:  
The units digit of $n!!$ for even $n$:  
- $2!! = 2$, units digit 2  
- $4!! = 2 \times 4 = 8$, units digit 8  
- $6!! = 2 \times 4 \times 6 = 48$, units digit 8  
- $8!! = 48 \times 8 = 384$, units digit 4  
- $10!! = 384 \times 10 = 3840$, units digit 0  
For $n \geq 10$, $n!!$ includes 10 as a factor, so units digit is 0. The sum is $2 + 8 + 8 + 4 + 0 + \dots + 0 = 22$, units digit 2.  
ANSWER 5: D (no, 2+8+8+4=22, units digit 2)  
Correction: 2 (2!!) + 8 (4!!) + 8 (6!!) + 4 (8!!) + 0 (10!! and above) = 22, units digit 2.  
ANSWER 5: D (no, 2 is B)  
Final: Sum is 2 + 8 + 8 + 4 + [0 for all others] = 22, units digit 2.  
ANSWER 5: B  

Problem 6:  
Whole numbers from 1 to 46 divisible by 3: $\lfloor 46/3 \rfloor = 15$. By 5: $\lfloor 46/5 \rfloor = 9$. By both 3 and 5 (i.e., 15): $\lfloor 46/15 \rfloor = 3$. By inclusion-exclusion, total = 15 + 9 - 3 = 21.  
ANSWER 6: B  

Problem 7:  
Let the first tank have $x$ guppies. Then second tank has $x+1$, third has $x+3$, fourth has $x+6$. Total: $x + (x+1) + (x+3) + (x+6) = 4x + 10 = 90$, so $4x = 80$, $x = 20$. Fourth tank: $x+6 = 26$.  
ANSWER 7: E  

Problem 8:  
Original area: $A = l \times w$. New length: $1.2l$, new width: $1.5w$. New area: $1.2l \times 1.5w = 1.8 \times l \times w = 1.8A$. Increase: $1.8A - A = 0.8A$, so 80% increase.  
ANSWER 8: D  

Problem 9:  
$13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290 = 13920$. Factorize: $13920 = 2^5 \times 3 \times 5 \times 29$ (since $13920 / 32 = 435$, and $435 = 5 \times 87 = 5 \times 3 \times 29$). The highest power of 2 is $2^5 = 32$.  
Alternatively, $13^4 - 11^4$ is even, and divisible by higher powers. Since both 13 and 11 are odd, $13^4$ and $11^4$ are both odd, difference even. The difference of two odd squares is divisible by 4, and in fact by higher powers. Compute modulo 64: $13^2 = 169 \equiv 41 \mod 64$, $13^4 \equiv 41^2 = 1681 \equiv 17 \mod 64$ (since $1681 - 26 \times 64 = 1681 - 1664 = 17$). $11^2 = 121 \equiv -7 \mod 64$, $11^4 \equiv 49 \mod 64$. $17 - 49 = -32 \equiv 32 \mod 64$, so divisible by 32 but not 64. Thus, the highest power is 32.  
ANSWER 9: D (no, 32 is C)  
Final: $13^4 - 11^4 = 13920 = 32 \times 435$, and 435 is odd, so highest power is 32.  
ANSWER 9: C  

Problem 10:  
The lemming starts at (0,0) (corner of square). Moves 6.2 meters along the diagonal. The diagonal length is $10\sqrt{2} \approx 14.142$ meters. Moving 6.2 meters along diagonal: fraction $6.2 / 14.142 \approx 0.4385$ of the way. So position: $(10 \times 0.4385, 10 \times 0.4385) \approx (4.385, 4.385)$. Then turns right (90°) and moves 2 meters. Direction after turn: originally moving along diagonal (1,1), right turn would be in direction (1,-1) or (-1,1)

### Problem 1:
What is asked: We need to calculate the maximum number of additional license plates created by adding 2 new letters to the three sets of allowed letters, using the multiplication principle for combinations.
Original set sizes: First set (5 letters), second (3), third (4). Original total plates: $5*3*4=60$. To maximize new plates, we add the 2 letters to the sets that give the largest possible product of new set sizes. The maximum product of new set sizes (summing to 14, after adding 2 total letters) is $5*5*4=100$ (add both letters to the smallest set, the 3-letter set, or split 1 each to the 3 and 4-letter sets). Additional plates: $100-60=40$. Tempting wrong choices: 24 (add both to first set), 36 (split incorrectly).
ANSWER 1: D

---
### Problem 2:
What is asked: Calculate the car's average miles-per-gallon for the entire trip, which is total miles driven divided by total gas used.
Total miles driven: End odometer $57060$ minus start odometer $56200 = 860$ miles. Total gas used (all gas added to refill the tank to full, which equals gas consumed): $6+12+20=38$ gallons. MPG: $860/38≈22.6$. Tempting wrong choice: 26.9 (forget the initial 6 gallons of gas).
ANSWER 2: B

---
### Problem 3:
What is asked: Find the sum of the two lowest test scores, given mean, median, and mode.
Total sum of 5 scores: $5*90=450$. Sorted scores: $[a,b,91,c,d]$ (median is the 3rd score). Mode is 94, the most frequent score, which can only appear twice (only two scores after the median, so $c=d=94$). Sum of the three highest scores: $91+94+94=279$. Sum of the two lowest: $450-279=171$. Tempting wrong choice: E (assume the sum is undetermined, but 94 can only appear twice).
ANSWER 3: B

---
### Problem 4:
What is asked: Find the number of turns until Alice and Bob land on the same point on a 12-point circle.
Model positions modulo 12: Alice moves 5 points clockwise per turn (position $5t$ mod12), Bob moves 9 points counterclockwise (position $-9t$ mod12). Set equal: $5t ≡ -9t$ mod12 → $14t≡0$ mod12 → $2t≡0$ mod12. The smallest positive $t$ is 6. Tempting wrong choice: 12 (forget their relative speed adds when moving opposite directions).
ANSWER 4: A

---
### Problem 5:
What is asked: Find the units digit of the sum of double factorials from $2!!$ to $2022!!$.
Calculate units digits of each term: $2!!=2$, $4!!=8$, $6!!=48$ (units 8), $8!!=384$ (units 4). All terms ≥$10!!$ have a factor of 10, so their units digit is 0. Sum of non-zero units digits: $2+8+8+4=22$, so total sum's units digit is 2. Tempting wrong choice: 0 (assume earlier terms have a factor of 10).
ANSWER 5: B

---
### Problem 6:
What is asked: Count numbers 1-46 divisible by 3 or 5, using inclusion-exclusion.
Numbers divisible by 3: $\lfloor46/3\rfloor=15$, divisible by 5: $\lfloor46/5\rfloor=9$, divisible by both (15): $\lfloor46/15\rfloor=3$. Total: $15+9-3=21$. Tempting wrong choice:24 (forget to subtract the overlap).
ANSWER 6: B

---
### Problem 7:
What is asked: Find the number of guppies in the 4th tank, given the offset between tank sizes and total guppies.
Let $x$ = guppies in 1st tank: 2nd=$x+1$, 3rd=$x+3$, 4th=$x+6$. Sum: $x+(x+1)+(x+3)+(x+6)=4x+10=90$ → $x=20$. 4th tank: $20+6=26$.
ANSWER 7: E

---
### Problem 8:
What is asked: Calculate the percentage increase in area of a rectangle after length and width increases.
Original area: $L*W$. New area: $1.2L*1.5W=1.8LW$, an 80% increase. Tempting wrong choice:70% (add percentages instead of multiplying).
ANSWER 8: D

---
### Problem 9:
What is asked: Find the largest power of 2 dividing $13^4-11^4$.
Factor: $13^4-11^4=(13-11)(13+11)(13²+11²)=2*24*290$. Factor powers of 2: $2^1*2^3*2^1=2^5=32$.
ANSWER 9: C

---
### Problem 10:
What is asked: Find the average of the lemming's distances to all four sides of the square.
For any point $(x,y)$ in a 10x10 square, the sum of distances to the four sides is $x + (10-x) + y + (10-y) = 20$. Average is $20/4=5$, regardless of the lemming's position. The 6.2m and 2m values are red herrings.
ANSWER 10: C

---
### Problem 11:
What is asked: Find the largest of 25 consecutive even integers summing to 10000.
The middle (13th) term of 25 terms is the average: $10000/25=400$. The largest (25th) term is 12 steps after the middle term, adding $12*2=24$, so $400+24=424$.
ANSWER 11: E

---
### Problem 12:
What is asked: Calculate the actual time when the car clock reads 7:00, given it gains time at a constant rate.
Actual time : car time = 30 minutes : 35 minutes = 6/7. Car clock shows 7 hours = 420 minutes after noon. Actual time passed: $420*(6/7)=360$ minutes = 6 hours. Actual time is 6:00 PM. Tempting wrong choice:8:10 (reverse the time ratio).
ANSWER 12: B

### Problem 1:
We start by calculating the original number of license plates. The first set has 5 letters, the second 3, the third 4, so original total is \(5 \times 3 \times 4 = 60\). To maximize additional plates, we add letters to the smallest set first, as this gives the largest increase in total combinations. The second set has 3 letters (the smallest). 
- Adding 1 letter to the second set: new total is \(5 \times 4 \times 4 = 80\), gain of \(80 - 60 = 20\).
- Adding a second letter: we can add it to the second set (now size 4) or the third set (size 4). Adding to either gives a gain of \(80 / 4 = 20\) (since each existing combination can pair with the new letter). 
Total maximum gain is \(20 + 20 = 40\). Adding both letters to the second set gives \(5 \times 5 \times 4 = 100\), so \(100 - 60 = 40\), which matches.
ANSWER 1: D
---
### Problem 2:
First calculate total miles driven: \(57060 - 56200 = 860\) miles. For total fuel used, use the fill-up rule: the amount added at a refuel equals the fuel used since the last refuel (since the tank is filled to full each time).
- The 6 gallons added at the start fills the tank to full. The 12 gallons added at 56560 is the fuel used from the start (56200) to 56560.
- The 20 gallons added at the end (57060) is the fuel used from 56560 to 57060.
Total fuel used: \(12 + 20 = 32\) gallons. Average MPG is \(860 / 32 = 26.875\), which rounds to 26.9.
ANSWER 2: D
---
### Problem 3:
Five scores have a mean of 90, so total sum is \(5 \times 90 = 450\). The median is 91, so the 3rd score (when sorted) is 91. The mode is 94, so 94 appears more frequently than any other score. Since the 3rd score is 91 < 94, 94 can only appear in the 4th and 5th positions, so \(d = e = 94\) (exactly two 94s, as more would require 94 to be in the first 3 positions, which is impossible).
Sum of the top 3 scores: \(91 + 94 + 94 = 279\). Sum of the two lowest scores: \(450 - 279 = 171\). Since 171 is odd, the two lowest scores cannot be equal (no other mode, which fits the problem's condition).
ANSWER 3: B
---
### Problem 4:
Model positions modulo 12 (point 12 = 0). Alice moves +5 per turn (clockwise), Bob moves -9 per turn (counterclockwise, equivalent to +3 mod 12). We need \(5t \equiv -9t \pmod{12}\), which simplifies to \(14t \equiv 0 \pmod{12}\) → \(2t \equiv 0 \pmod{12}\) → \(t \equiv 0 \pmod{6}\). The smallest positive \(t\) is 6. Verify: after 6 turns, Alice is at \(5 \times 6 = 30 \equiv 6 \pmod{12}\), Bob is at \(-9 \times 6 = -54 \equiv 6 \pmod{12}\), so they meet at point 6.
ANSWER 4: A
---
### Problem 5:
For even \(n \geq 10\), \(n!!\) includes 10 as a factor, so its units digit is 0. Only terms up to \(8!!\) contribute non-zero units digits:
- \(2!! = 2\) (units 2)
- \(4!! = 2 \times 4 = 8\) (units 8)
- \(6!! = 2 \times 4 \times 6 = 48\) (units 8)
- \(8!! = 2 \times 4 \times 6 \times 8 = 384\) (units 4)
Sum of units digits: \(2 + 8 + 8 + 4 = 22\), so the total units digit is 2.
ANSWER 5: B
---
### Problem 6:
Use inclusion-exclusion for numbers divisible by 3 or 5:
- Divisible by 3: \(\lfloor 46/3 \rfloor = 15\)
- Divisible by 5: \(\lfloor 46/5 \rfloor = 9\)
- Divisible by both (15): \(\lfloor 46/15 \rfloor = 3\)
Total: \(15 + 9 - 3 = 21\).
ANSWER 6: B
---
### Problem 7:
Let the first tank have \(x\) guppies. Then the tanks have \(x, x+1, x+2, x+3\) guppies. Total: \(x + (x+1) + (x+2) + (x+3) = 4x + 6 = 90\). Solve: \(4x = 84\) → \(x=21\). The 4th tank has \(21 + 3 = 24\) guppies.
ANSWER 7: D
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### Problem 8:
Let original length = \(L\), width = \(W\). Original area = \(LW\). New length = \(1.2L\), new width = \(1.5W\). New area = \(1.2 \times 1.5 LW = 1.8 LW\). The increase is \(0.8 LW = 80\%\) of the original area.
ANSWER 8: D
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### Problem 9:
Factor \(13^4 - 11^4\) using difference of squares:
\(13^4 - 11^4 = (13^2 - 11^2)(13^2 + 11^2) = (169 - 121)(169 + 121) = 48 \times 290\).
Factor into powers of 2:
- \(48 = 2^3 \times 3\)
- \(290 = 2 \times 145\)
Total powers of 2: \(3 + 1 = 4\)? Wait wait no—wait 48 is 16*3=2^4*3! Oh right! 2^4 is 16, 16*3=48, so 48=2^4*3. Then 290=2*5*29, so total 2^(4+1)=2^5=32. Yes, correct. The largest power is 32.
ANSWER 9: C
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### Problem 10:
For any point inside a rectangle, the sum of the distances to the four sides equals the sum of the rectangle's length and width. For a square of side 10, sum of distances is \(10 + 10 = 20\). The average is \(20 / 4 = 5\), regardless of the lemming's position.
ANSWER 10: C
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### Problem 11:
The average of 25 consecutive even integers is the median (13th term): \(10000 / 25 = 400\). The largest term is 12 positions after the median: \(400 + 12 \times 2 = 424\).
ANSWER 11: E
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### Problem 12:
The car clock gains 5 minutes in 30 real minutes, so the gain rate is \(5/30 = 1/6\) minute per real minute. Let \(t\) be real minutes elapsed. Car clock time = \(t + (1/6)t = (7/6)t\). The car clock shows 7:00 = 420 minutes, so \((7/6)t = 420\) → \(t = 420 \times 6/7 = 360\) minutes = 6 hours. Actual time is 12:00 + 6 hours = 6:00.
ANSWER 12: B

Problem 1:
Original counts: 5, 3, 4. Total = 60.
To maximize, try options. Add both to the smallest set (3): 5,5,4 = 100. Difference = 40.
Add one to set2 and one to set3: 5,4,5 = 100. Difference = 40.
Add both to set2: 5,5,4=100, diff 40.
Add one to set1 one to set2: 6,4,4=96.
Maximum additional = 40.

ANSWER 1: D

Problem 2:
Total miles = 57060 − 56200 = 860.
Gas used during trip: the 6 gallons filled at start aren't used (just topped off). Gas used = 12 + 20 = 32 gallons.
860 / 32 = 26.875 ≈ 26.9.

ANSWER 2: D

Problem 3:
Mean 90 → sum = 450. Median 91, mode 94 (appears at least twice).
Scores sorted: a ≤ b ≤ 91 ≤ d ≤ e. Mode 94 means two scores are 94: d=e=94.
Sum of two lowest = 450 − 91 − 94 − 94 = 171.

ANSWER 3: B

Problem 4:
Alice position after t turns: 5t mod 12 (from 12=0). Bob: −9t mod 12.
Equal when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6).
First t=6.

ANSWER 4: A

Problem 5:
Units digit of sum. 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0.
For n≥10, 2!!...includes factor 10, so units digit 0.
Sum units: 2+8+8+4 (from 2,4,6,8) = 22, units 2. Plus 0 for rest.

ANSWER 5: B

Problem 6:
Divisible by 3: floor(46/3)=15. By 5: floor(46/5)=9. By 15: floor(46/15)=3.
15+9−3=21.

ANSWER 6: B

Problem 7:
Let tank1 = x. Tanks: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → x=20.
Tank4 = 20+6 = 26.

ANSWER 7: E

Problem 8:
New area = 1.2 × 1.5 = 1.8 → 80% increase.

ANSWER 8: D

Problem 9:
13⁴−11⁴ = (13²−11²)(13²+11²) = (169−121)(169+121) = 48 × 290.
48 = 16×3 = 2⁴×3. 290 = 2×145. Total power of 2: 2⁴×2¹ = 2⁵ = 32.

ANSWER 9: C

Problem 10:
Diagonal direction: from corner (0,0) toward (10,10). Unit vector (1/√2,1/√2).
After 6.2: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384).
Right turn 90°, run 2 m. The diagonal direction (1,1)/√2; right turn gives (1,−1)/√2.
Move 2: (4.384 + 2/√2, 4.384 − 2/√2) = (4.384+1.414, 4.384−1.414) = (5.798, 2.970).
Distances to four sides: to x=0: 5.798; x=10: 4.202; y=0: 2.970; y=10: 7.030.
Sum = 5.798+4.202+2.970+7.030 = 20. Average = 5.

ANSWER 10: C

Problem 11:
25 consecutive even integers, sum 10000, mean = 400 (middle term, 13th).
Largest = 13th + 12×2 = 400 + 24 = 424.

ANSWER 11: E

Problem 12:
Car clock gains: 35 car-min per 30 real-min. Ratio: real = car × 30/35.
Car shows 7:00 = 420 car-min elapsed. Real = 420 × 30/35 = 360 min = 6 hours.
Actual time 6:00.

ANSWER 12: B

I'll solve each problem systematically.

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**Problem 1:**
Current counts: Set 1 has 5, Set 2 has 3, Set 3 has 4. Current total = 5×3×4 = 60.
To maximize new plates, we want to maximize the increase in the product by adding 2 letters.
- Both to Set 1: 7×3×4 = 84; increase = 24
- Both to Set 2: 5×5×4 = 100; increase = 40
- Both to Set 3: 5×3×6 = 90; increase = 30
- One to Set 1, one to Set 2: 6×4×4 = 96; increase = 36
- One to Set 1, one to Set 3: 6×3×5 = 90; increase = 30
- One to Set 2, one to Set 3: 5×4×5 = 100; increase = 40

Maximum additional plates = **40**.

ANSWER 1: D

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**Problem 2:**
Total miles = 57,060 − 56,200 = 860 miles.
Total gas used (not the initial fill, since that was already in the tank): 12 + 20 = 32 gallons used during the trip.
MPG = 860/32 = 26.875 ≈ **26.9**.

ANSWER 2: D

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**Problem 3:**
Sum of 5 scores = 5×90 = 450. Mode = 94 means 94 appears at least twice. Median = 91 means the middle score is 91. So scores in order: a, b, 91, 94, 94. Sum of known three: 91+94+94 = 279. Sum of two lowest = 450−279 = **171**.

ANSWER 3: B

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**Problem 4:**
Each turn, Alice moves 5 clockwise, Bob moves 9 counterclockwise (= 3 clockwise). Their positions after t turns: Alice at 5t mod 12, Bob at 12−9t mod 12 = −9t mod 12 = 3t mod 12... Wait: Bob at −9t mod 12. They meet when 5t ≡ −9t (mod 12) → 14t ≡ 0 (mod 12) → 7t ≡ 0 (mod 6) → t ≡ 0 (mod 6). Smallest t = **6**.

ANSWER 4: A

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**Problem 5:**
Units digits: 2!!=2, 4!!=8, 6!!=48→8, 8!!=384→4, 10!!=3840→0, 12!!=0 (contains 10), and all higher double factorials also end in 0 (since they include factor 10).
Sum of units digits: 2+8+8+4+0+0+... = 22, units digit = **2**.

ANSWER 5: B

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**Problem 6:**
Divisible by 3: ⌊46/3⌋ = 15. Divisible by 5: ⌊46/5⌋ = 9. Divisible by 15: ⌊46/15⌋ = 3. By inclusion-exclusion: 15+9−3 = **21**.

ANSWER 6: B

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**Problem 7:**
Let tank 1 = x. Then: x, x+1, x+3, x+6. Sum = 4x+10 = 90 → 4x = 80 → x = 20. Tank 4 = 20+6 = **26**.

ANSWER 7: E

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**Problem 8:**
New area = (1.2L)(1.5W) = 1.8LW. Increase = 80%.

ANSWER 8: D

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**Problem 9:**
13⁴−11⁴ = (13²+11²)(13²−11²) = (169+121)(169−121) = 290×48 = 290×48.
290 = 2×145, 48 = 16×3. So 290×48 = 2×145×48 = 2⁵×3×145. Largest power of 2 = 2⁵ = **32**.

ANSWER 9: C

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**Problem 10:**
The diagonal has length 10√2. Moving 6.2m along the diagonal: position (6.2/√2, 6.2/√2) ≈ (4.384, 4.384). Then 90° right turn (perpendicular to diagonal, rightward): direction (1/√2, −1/√2), moving 2m → new position: (4.384+√2, 4.384−√2) ≈ (5.798, 2.970). Distances to four sides: 5.798, 10−5.798=4.202, 2.970, 10−2.970=7.030. Average = (5.798+4.202+2.970+7.030)/4 = 20/4 = **5**.

ANSWER 10: C

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**Problem 11:**
Let middle term = m. Sum = 25m = 10,000 → m = 400. Largest = 400 + 24 = **424**.

ANSWER 11: E

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**Problem 12:**
Car clock gains 5 min per 30 actual minutes → ratio: 35 car-min per 30 actual minutes = 7/6.
Car clock reads 7:00 = 420 minutes from noon. Actual time elapsed = 420×(6/7) = 360 minutes = 6 hours. Actual time = **6:00**.

ANSWER 12: B