🔬 Model Lab

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles 

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

---

**Problem 1:**

**What is being asked:** Find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9 exactly once each.

**Setup:** We want to minimize the sum of two 3-digit numbers. To minimize a sum, we want to minimize the hundreds places first, then tens, then units.

**Strategy:** Put the two smallest digits (4 and 5) in the hundreds places. Then put the next two smallest (6 and 7) in the tens places. Put the largest (8 and 9) in the units places.

This gives us: 468 + 579 = 1047

Let me verify this is minimal by trying 469 + 578 = 1047 (same)

Let me try 467 + 589 = 1056 (larger)

Let me try 458 + 679 = 1137 (larger)

The arrangement 468 + 579 = 1047 is optimal.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many half-dozens of eggs should Walter buy to make enough cookies for 108 students eating 2 cookies each?

**Setup:**
- Total cookies needed: 108 × 2 = 216 cookies
- Each pan makes 15 cookies
- Number of pans needed: 216 ÷ 15 = 14.4, so 15 pans (round up)
- Eggs per pan: 2 eggs
- Total eggs needed: 15 × 2 = 30 eggs
- Eggs per half-dozen: 6 eggs
- Half-dozens needed: 30 ÷ 6 = 5

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** If 6 pepperoni circles fit across the diameter of a 12-inch pizza, and 24 are placed without overlap, what fraction of pizza is covered?

**Setup:**
- Pizza diameter: 12 inches, so radius = 6 inches
- If 6 pepperoni circles fit across the diameter: diameter of one pepperoni = 12 ÷ 6 = 2 inches
- Radius of one pepperoni: 1 inch
- Area of one pepperoni: π(1)² = π square inches
- Total area of 24 pepperoni: 24π square inches
- Area of pizza: π(6)² = 36π square inches
- Fraction covered: 24π ÷ 36π = 24/36 = 2/3

ANSWER 3: B

---

**Problem 4:**

**What is being asked:** How many games in a league with 2 divisions of 6 teams each, where teams play division rivals twice and other division teams once?

**Setup:**
- Each division: 6 teams
- Games within a division: Each team plays 5 other teams twice = 10 games per team
  - Total division games: (6 × 10) ÷ 2 = 30 games per division
  - Both divisions: 2 × 30 = 60 games
- Games between divisions: Each team plays all 6 teams in other division once
  - Total: 6 × 6 = 36 games
- Total games: 60 + 36 = 96 games

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** Total widgets sold after 20 days, where day n has sales of 2n - 1 widgets.

**Setup:**
- Day 1: 1 widget
- Day 2: 3 widgets
- Day 3: 5 widgets
- Pattern: Day n sells (2n - 1) widgets

This is the sum of the first 20 odd numbers: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers = n²

For n = 20: 20² = 400 widgets

ANSWER 5: D

---

**Problem 6:**

**What is being asked:** How many minutes to download a 4.2-megabyte song at 56 kilobits per second?

**Setup:**
- File size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits
- Speed: 56 kilobits per second
- Time in seconds: 33,600 ÷ 56 = 600 seconds
- Time in minutes: 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

**Problem 7:**

**What is being asked:** Rank the friends given exactly one statement is true among: I. Bill is oldest, II. Amy is not oldest, III. Celine is not youngest.

**Testing each case:**

**Case: Only I is true** (Bill oldest)
- II is false: Amy IS oldest (contradiction with I)
- This doesn't work.

**Case: Only II is true** (Amy is not oldest)
- I is false: Bill is not oldest
- III is false: Celine IS youngest
- So order is: Bill oldest, then Amy, then Celine youngest
- Check: Amy not oldest ✓, Bill not oldest ✓, Celine is youngest ✓

**Case: Only III is true** (Celine is not youngest)
- I is false: Bill not oldest
- II is false: Amy IS oldest
- So Amy oldest, then Celine (not youngest), then Bill youngest
- Check: Celine not youngest ✓, Bill not oldest ✓, Amy is oldest ✓

Comparing cases: Only case II works (Bill, Amy, Celine matches choice A).

ANSWER 7: A

---

**Problem 8:**

**What is being asked:** How many 4-digit numbers greater than 1000 use the four digits of 2012 (digits: 0, 1, 2, 2)?

**Setup:** We have digits {0, 1, 2, 2}. The first digit cannot be 0.

**First digit choices:** 1 or 2

**If first digit is 1:** Arrange {0, 2, 2}: positions can be filled with 0, 2, 2 in 3!/2! = 3 ways
- 1022, 1202, 1220

**If first digit is 2:** Arrange {0, 1, 2}: positions can be filled in 3! = 6 ways
- 2012, 2021, 2102, 2120, 2201, 2210

Total: 3 + 6 = 9 numbers

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Probability that a 3-digit number formed from drawing 3 of {1, 2, 3, 4} is divisible by 3.

**Setup:** A number is divisible by 3 if the sum of digits is divisible by 3.

Total 3-digit numbers: P(4,3) = 4 × 3 × 2 = 24

**Possible digit sets and their sums:**
- {1,2,3}: sum = 6 (divisible by 3) → 3! = 6 numbers
- {1,2,4}: sum = 7 (not divisible by 3)
- {1,3,4}: sum = 8 (not divisible by 3)
- {2,3,4}: sum = 9 (divisible by 3) → 3! = 6 numbers

Total divisible by 3: 6 + 6 = 12 numbers

Probability: 12/24 = 1/2

ANSWER 9: C

---

**Problem 10:**

**What is being asked:** Which runner finished fourth (given various time relationships)?

**Setup:** Let P = Pedro's time (reference)
- Nico: P + 11
- Olympia: P - 3
- Melina: P - 3 + 2 = P - 1
- Luke: P - 3 - 6 = P - 9

**Order from fastest (smallest time) to slowest:**
1. Luke: P - 9
2. Olympia: P - 3
3. Melina: P - 1
4. Pedro: P
5. Nico: P + 11

Pedro finished fourth.

ANSWER 10: E

---

**Problem 11:**

**What is being asked:** Find n where 20 quarters + 10 dimes = 10 quarters + n dimes.

**Setup:**
- Left side: 20(0.25) + 10(0.10) = 5.00 + 1.00 = 6.00 dollars
- Right side: 10(0.25) + n(0.10) = 2.50 + 0.10n dollars

Setting equal: 6.00 = 2.50 + 0.10n
- 3.50 = 0.10n
- n = 35

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** In how many months will Brent and Gretel have the same number of goldfish?

**Setup:**
- Brent starts with 4, quadruples each month: 4 × 4^m
- Gretel starts with 128, doubles each month: 128 × 2^m

Setting equal: 4 × 4^m

Problem 1:

We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so that their sum is as small as possible.

To make the sum as small as possible, we want the smallest digits in the hundreds places, the next smallest in the tens places, and the largest digits in the ones places, because hundreds count the most, then tens, then ones.

So put:
- 4 and 5 in the hundreds places
- 6 and 7 in the tens places
- 8 and 9 in the ones places

Then the sum is:

\[
400+500+60+70+8+9=1047
\]

So the smallest possible sum is 1047.

Now rule out the choices:
- 947 is too small to be possible because the two hundreds digits must already total at least 900.
- 1037 is close, but not as small as 1047.
- 1056, 1245 are larger.

ANSWER 1: C

Problem 2:

108 students eat an average of 2 cookies each, so the total number of cookies needed is:

\[
108 \times 2 = 216
\]

Each pan makes 15 cookies. Since only full recipes are made, we need enough pans to make at least 216 cookies.

\[
216 \div 15 = 14.4
\]

So Walter must make 15 pans.

Each pan uses 2 eggs, so the number of eggs needed is:

\[
15 \times 2 = 30
\]

Walter buys eggs by the half-dozen, and one half-dozen is 6 eggs. Thus:

\[
30 \div 6 = 5
\]

So he should buy 5 half-dozens.

Check choices:
- 1, 2 are far too few
- 7 or 15 half-dozens are too many
- 5 is correct

ANSWER 2: C

Problem 3:

We are told 6 pepperoni circles exactly fit across the diameter of a 12-inch pizza.

That means each pepperoni has diameter:

\[
12 \div 6 = 2 \text{ inches}
\]

So each pepperoni has radius 1 inch.

Area of one pepperoni:

\[
\pi(1)^2 = \pi
\]

Area of 24 pepperoni circles:

\[
24\pi
\]

Now area of the pizza (radius 6 inches):

\[
\pi(6)^2 = 36\pi
\]

So the fraction covered is:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

Thus the answer is \(\frac{2}{3}\).

Checking the choices, that is B.

ANSWER 3: B

Problem 4:

There are 2 divisions with 6 teams each.

First, count games within one division:
Each team plays every other team in its division twice.

Number of pairs of teams in one division:

\[
\binom{6}{2}=15
\]

Each pair plays 2 games, so games within one division:

\[
15 \times 2 = 30
\]

There are 2 divisions, so total within-division games:

\[
2 \times 30 = 60
\]

Now count games between divisions:
Each of the 6 teams in one division plays each of the 6 teams in the other division once:

\[
6 \times 6 = 36
\]

Total games:

\[
60 + 36 = 96
\]

So the answer is 96.

ANSWER 4: B

Problem 5:

Janabel sold:
- Day 1: 1
- Day 2: 3
- Day 3: 5

This is an arithmetic sequence increasing by 2 each day.

The number sold on day \(n\) is:

\[
1, 3, 5, 7, \dots
\]

So on day 20, she sold:

\[
1 + 2(20-1) = 39
\]

But the question asks for the total sold after 20 days, so we need the sum of the first 20 odd numbers.

A known fact is:

\[
1+3+5+\cdots+(2n-1)=n^2
\]

So after 20 days:

\[
20^2 = 400
\]

Check choices:
- 39 is only day 20
- 400 is the total after 20 days

ANSWER 5: D

Problem 6:

A 4.2-megabyte song is downloaded at 56 kilobits per second.

Given:
- 1 megabyte = 8000 kilobits

So the song size in kilobits is:

\[
4.2 \times 8000 = 33600 \text{ kilobits}
\]

At 56 kilobits per second, time in seconds is:

\[
33600 \div 56 = 600 \text{ seconds}
\]

Convert to minutes:

\[
600 \div 60 = 10 \text{ minutes}
\]

So the answer is 10 minutes.

ANSWER 6: B

Problem 7:

We have three friends with different ages: Amy, Bill, and Celine.

Exactly one of these statements is true:
I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

We test the possible rankings.

Suppose Bill is the oldest. Then statement I is true. Since exactly one statement is true, statements II and III must be false.

- II false means Amy is the oldest, impossible if Bill is oldest.
So Bill cannot be oldest.

Now suppose Amy is oldest. Then statement II (“Amy is not the oldest”) is false.

To make exactly one statement true, we need either I or III true, but not both.

If Celine is youngest, then III is false, so only one true is impossible because I is also false.  
If Celine is not youngest, then III is true and I is false, so exactly one statement is true.

Thus:
- Amy is oldest
- Celine is not youngest
Since ages are different, the only arrangement is:

\[
\text{Amy, Celine, Bill}
\]

Check:
- I: Bill is oldest — false
- II: Amy is not oldest — false
- III: Celine is not youngest — true

Exactly one statement is true.

ANSWER 7: E

Problem 8:

We need 4-digit numbers greater than 1000 using the digits of 2012.

The digits are 2, 0, 1, 2. Since there are two 2s, we count distinct arrangements.

We want 4-digit numbers greater than 1000, so the first digit cannot be 0.

Count all distinct permutations of 0,1,2,2:

\[
\frac{4!}{2!}=12
\]

Now subtract those starting with 0. If 0 is first, the remaining digits 1,2,2 can be arranged in:

\[
\frac{3!}{2!}=3
\]

So the number greater than 1000 is:

\[
12-3=9
\]

Thus the answer is 9.

ANSWER 8: D

Problem 9:

We have digits 1, 2, 3, 4. We draw 3 without replacement and form a 3-digit number.

We want the probability the number is a multiple of 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now check all 3-digit selections from {1,2,3,4}:

- 1 + 2 + 3 = 6, divisible by 3
- 1 + 2 + 4 = 7, not divisible
- 1 + 3 + 4 = 8, not divisible
- 2 + 3 + 4 = 9, divisible by 3

So 2 of the 4 possible sets of 3 digits work.

Each set of 3 digits can be arranged in \(3! = 6\) ways, and each arrangement is equally likely. Since the number of favorable arrangements and total arrangements both get multiplied by 6, the probability is simply:

\[
\frac{2}{4}=\frac{1}{2}
\]

So the answer is \(\frac{1}{2}\).

ANSWER 9: C

Problem 10:

Let Pedro’s finishing time be \(P\).

Then:
- Olympia finished 3 minutes behind Pedro, so \(O = P+3\)
- Olympia finished 2 minutes ahead of Melina, so \(M = O+2 = P+5\)
- Nico finished 11 minutes behind Pedro, so \(N = P+11\)
- Olympia finished 6 minutes ahe

Problem 1:  
We are adding two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once total. We want the **smallest possible sum**.  
To make the sum as small as possible:
- Put the **smallest digits in the highest place values** (hundreds place), because hundreds place affects the number most.
- Also, for two 3-digit numbers, make both numbers as small as possible.

The two 3-digit numbers are of the form:
\[
\overline{abc}+\overline{def}
\]
with digits \(\{4,5,6,7,8,9\}\).

**Step 1: Choose the hundreds digits (smallest two digits).**  
Smallest digits are 4 and 5.  
So the hundreds places should be 4 and 5:
\[
\overline{4..}+\overline{5..}
\]

**Step 2: Choose remaining digits to minimize the overall sum.**  
After using 4 and 5, the remaining digits are \(\{6,7,8,9\}\).  
To minimize the numbers, the next smallest digits go to tens places, then ones places. The best arrangement is:
- tens digits: 6 and 7
- ones digits: 8 and 9

So the smallest possible pair is:
\[
468 + 579 = 1047
\]
(Any swap that makes one number larger without making the other smaller enough will increase the total.)

Check quickly:
- If we tried \(468+579\), sum \(=1047\).
- Choice C is 1047.

**Answer check:** choices less than 1047: 947, 1037. Those would require even smaller hundreds/tens placement than using 4 and 5 as hundreds digits, impossible.

ANSWER 1: C


Problem 2:  
108 students eat an average of 2 cookies each, so total cookies eaten:
\[
108 \cdot 2 = 216 \text{ cookies}
\]

Each full recipe pan makes 15 cookies. Number of full pans needed:
\[
\frac{216}{15} = 14.4
\]
So we must make **15 full pans** (since only full recipes are made).

Eggs: 2 eggs per pan, so eggs needed:
\[
15 \cdot 2 = 30 \text{ eggs}
\]

Walter buys eggs by the half-dozen = 6 eggs each. Number of half-dozens:
\[
\frac{30}{6} = 5
\]

ANSWER 2: C


Problem 3:  
Across a diameter of 12 inches, 6 circles fit exactly.  
So each circle’s diameter is:
\[
\frac{12}{6} = 2 \text{ inches}
\]
Thus each pepperoni circle has radius \(1\) and area:
\[
\pi \cdot 1^2 = \pi
\]

The pizza’s radius is \(6\) inches, so pizza area:
\[
\pi \cdot 6^2 = 36\pi
\]

If 24 circles are placed without overlap, total pepperoni area:
\[
24\pi
\]

Fraction covered:
\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}
\]

So the fraction is \(\boxed{\frac{2}{3}}\), which is choice B.

ANSWER 3: B


Problem 4:  
There are 6 teams in each division, so 12 teams total.

Each team plays:
- Other teams in its **own** division: 2 times each.
- Teams in the **other** division: 1 time each.

Fix a team in one division.

**Games within its division:**  
There are 5 other teams in its division, played twice:
\[
5 \cdot 2 = 10 \text{ games per team}
\]

**Games against other division:**  
There are 6 teams in the other division, played once:
\[
6 \cdot 1 = 6 \text{ games per team}
\]

So each team plays:
\[
10+6=16 \text{ games}
\]

Total “team-games” across all 12 teams:
\[
12 \cdot 16 = 192
\]
But each actual game is counted twice (once for each team), so divide by 2:
\[
\frac{192}{2}=96
\]

ANSWER 4: B


Problem 5:  
The number sold each day forms an arithmetic sequence:
- Day 1: 1  
- Day 2: 3  
- Day 3: 5  
Difference each day is \(+2\).

So on day \(n\):
\[
a_n = 1 + (n-1)\cdot 2 = 2n-1
\]

We want total after 20 days:
\[
S_{20} = \frac{20}{2}(a_1+a_{20})
\]
Compute \(a_{20}\):
\[
a_{20} = 2(20)-1 = 39
\]
So:
\[
S_{20} = 10(1+39)=10\cdot 40=400
\]

ANSWER 5: D


Problem 6:  
We need time to download a 4.2-megabyte song at 56 kilobits/sec.

Convert megabytes to kilobits.
- 8000 kilobits in 1 megabyte.

So:
\[
4.2 \text{ MB} = 4.2 \cdot 8000 = 33600 \text{ kilobits}
\]

Speed is 56 kilobits per second, so time in seconds:
\[
\frac{33600}{56} = 600 \text{ seconds}
\]

Convert to minutes:
\[
\frac{600}{60}=10 \text{ minutes}
\]

ANSWER 6: B


Problem 7:  
We have three people with different ages. Exactly one statement is true:

I. Bill is the oldest.  
II. Amy is not the oldest.  
III. Celine is not the youngest.

Let’s test the possible rankings (oldest → youngest).  

**Option A: Bill, Amy, Celine**  
- Bill is oldest: I is true.  
- Amy is not oldest: II is true.  
- Celine is not youngest: but Celine is youngest, so III is false.  
That gives **two truths**, not allowed.

**Option B: Amy, Bill, Celine**  
- I: Bill oldest? No (Amy oldest) → false  
- II: Amy not oldest? Amy is oldest → II false  
- III: Celine not youngest? Celine is youngest → false  
That gives **zero truths**, not allowed.

**Option C: Celine, Amy, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is middle, so yes → true  
- III: Celine not youngest? Celine is oldest, so indeed Celine is not youngest → true  
Two truths, not allowed.

**Option D: Celine, Bill, Amy**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is youngest, so Amy is not oldest → true  
- III: Celine not youngest? Celine is oldest, so not youngest → true  
Two truths, not allowed.

**Option E: Amy, Celine, Bill**  
- I: Bill oldest? No → false  
- II: Amy not oldest? Amy is oldest → false  
- III: Celine not youngest? Celine is middle, so not youngest → true  
Exactly one statement true (III). Works.

So ranking oldest → youngest is Amy, Celine, Bill.

ANSWER 7: E


Problem 8:  
How many 4-digit numbers > 1000 can be formed using digits of 2012?  
Digits available: 2,0,1,2. Note digit 2 repeats, so there are 4 digits with repetition.

Total distinct permutations of 4 digits:
\[
\frac{4!}{2!}=\frac{24}{2}=12
\]

But we need numbers **greater than 1000**, meaning the thousands digit cannot be 0.

Count permutations where thousands digit is 0:
- Fix thousands digit = 0.
- Remaining digits are 2,1,2 (three digits with 2 repeated):
\[
\frac{3!}{2!}=\frac{6}{2}=3
\]

So valid numbers:
\[
12-3=9
\]

ANSWER 8: D


Problem 9:  
We draw 3 digits without replacement from \(\{1,2,3,4\}\). Total equally likely 3-digit numbers (order matters):
\[
4\cdot 3\cdot 2 = 24
\]

We want probability the 3-digit number i

Problem 1:
To minimize the sum of two 3-digit numbers formed by the digits {4, 5, 6, 7, 8, 9}, we must place the smallest digits in the hundreds place. Let the numbers be $ABC$ and $DEF$. The sum is $(100A+10B+C) + (100D+10E+F) = 100(A+D) + 10(B+E) + (C+F)$. To minimize this, we assign the smallest digits {4, 5} to the hundreds place ($A, D$). The next smallest digits {6, 7} go to the tens place ($B, E$), and the largest {8, 9} go to the units place ($C, F$).
Sum = $100(4+5) + 10(6+7) + (8+9) = 900 + 130 + 17 = 1047$.
ANSWER 1: C

Problem 2:
Total cookies needed = $108 \times 2 = 216$. Each pan makes 15 cookies. Number of pans = $216 / 15 = 14.4$. Since only full recipes (pans) are made, he must make 15 pans. Each pan uses 2 eggs, so he needs $15 \times 2 = 30$ eggs. Since eggs are sold by the half-dozen (6 eggs), he needs $30 / 6 = 5$ half-dozens.
ANSWER 2: C

Problem 3:
The pizza has a diameter of 12 inches, so its radius $r = 6$ inches. Area of pizza = $\pi(6^2) = 36\pi$. Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has a diameter of 2 inches and a radius of 1 inch. Area of one pepperoni = $\pi(1^2) = \pi$. Total area of 24 pepperonis = $24\pi$. The fraction covered = $24\pi / 36\pi = 24/36 = 2/3$.
ANSWER 3: B

Problem 4:
Each division has 6 teams. Within a division, each team plays the other 5 teams twice: $6 \times 5 = 30$ games. Across divisions, each of the 6 teams in Div A plays each of the 6 teams in Div B once: $6 \times 6 = 36$ games. Total games = $30 + 36 = 66$. Wait, let's re-calculate: Within Div A: $6 \times 5 / 2 = 15$ pairs, each plays twice = 30 games. Within Div B: 30 games. Between Divs: $6 \times 6 = 36$ games. Total = $30 + 30 + 36 = 96$.
ANSWER 4: B

Problem 5:
This is an arithmetic progression: $1, 3, 5, \dots$. The $n$-th term is $a_n = 1 + (n-1)2 = 2n - 1$. The sum of the first 20 terms is $S_{20} = \frac{n}{2}(a_1 + a_{20}) = \frac{20}{2}(1 + (2(20)-1)) = 10(1 + 39) = 10(40) = 400$.
ANSWER 5: D

Problem 6:
Total kilobits = $4.2 \text{ MB} \times 8000 \text{ kilobits/MB} = 33600 \text{ kilobits}$. Speed = 56 kilobits/sec. Time in seconds = $33600 / 56 = 600$ seconds. Time in minutes = $600 / 60 = 10$ minutes.
ANSWER 6: B

Problem 7:
Let ages be $A, B, C$. Statements: (I) $B$ is oldest, (II) $A$ is not oldest, (III) $C$ is not youngest. If (I) is true, then (II) and (III) must be false. If (II) is false, $A$ is oldest (Contradiction). If (II) is true, then (I) and (III) are false. If (I) is false, $B$ is not oldest. If (III) is false, $C$ is youngest. If $A$ is oldest and $C$ is youngest, then $B$ is middle. Order: $A, B, C$. Check: (I) $B$ oldest (False), (II) $A$ not oldest (False - wait). Let's test (III) true: $C$ is oldest or middle, $A$ is oldest, $B$ is youngest. If $C$ is oldest, $A$ is middle, $B$ is youngest: (I) False, (II) True, (III) True (Contradiction). If $A$ is oldest, $C$ is middle, $B$ is youngest: (I) False, (II) False, (III) True. This works. Order: $A, C, B$.
ANSWER 7: E

Problem 8:
Digits: {2, 0, 1, 2}. Permutations: 2012, 2021, 2102, 2120, 2201, 2210, 0122 (not 4-digit), 0212 (not 4-digit), 0221 (not 4-digit), 1022, 1202, 1220. Total 4-digit numbers: 2012, 2021, 2102, 2120, 2201, 2210, 1022, 1202, 1220. That is 9 numbers.
ANSWER 8: D

Problem 9:
Total permutations of 3 digits from {1, 2, 3, 4} is $4 \times 3 \times 2 = 24$. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Possible sets of 3 digits: {1,2,3} (sum 6, yes), {1,2,4} (sum 7, no), {1,3,4} (sum 8, no), {2,3,4} (sum 9, yes). Each set has $3! = 6$ permutations. Total successful outcomes = $6 + 6 = 12$. Probability = $12/24 = 1/2$.
ANSWER 9: C

Problem 10:
Let Pedro be at time 0. Nico = +11. Olympia = +3. Melina = $3-2 = +1$. Luke = $3+6 = +9$. Order (earliest to latest): Pedro (0), Melina (1), Olympia (3), Luke (9), Nico (11). Fourth is Luke.
ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes = $20(25) + 10(10) = 500 + 100 = 600$. Value of 10 quarters and $n$ dimes = $10(25) + 10n = 250 + 10n$. Set equal: $600 = 250 + 10n \implies 350 = 10n \implies n = 35$.
ANSWER 11: D

Problem 12:
Brent: $B(t) = 4 \times 4^t = 4^{t+1}$. Gretel: $G(t) = 128 \times 2^t = 2^7 \times 2^t = 2^{t+7}$. Set $4^{t+1} = 2^{t+7} \implies (2^2)^{t+1} = 2^{t+7} \implies 2^{2t+2} = 2^{t+7}$. Thus $2t + 2 = t + 7 \implies t = 5$.
ANSWER 12: B

Problem 1:
To minimize the sum of two 3-digit numbers formed from the digits 4 through 9 used exactly once, the hundreds digits must be 4 and 5 (the smallest possible pair). The remaining digits are then 6, 7, 8, and 9. The arrangements that minimize the overall sum are those that also keep the tens digits as small as possible while respecting the digit constraint:
468 + 579 = 1047,
469 + 578 = 1047,
478 + 569 = 1047,
479 + 568 = 1047.
All other pairings of hundreds digits produce strictly larger sums (e.g., 467 + 589 = 1056). Thus the minimal sum is 1047.
ANSWER 1: C
Problem 2:
108 students each eating an average of 2 cookies requires a total of \(108 \times 2 = 216\) cookies. Each pan yields 15 cookies, so the number of full pans needed is the smallest integer at least \(216/15 = 14.4\), which is 15 pans. Each pan uses 2 eggs, so 15 pans require 30 eggs. Eggs are purchased by the half-dozen (6 eggs), and \(30/6 = 5\) half-dozens are therefore required.
ANSWER 2: C
Problem 3:
A 12-inch pizza has diameter 12 and radius 6, so its area is \(\pi \times 6^2 = 36\pi\). Six pepperoni circles fit exactly across the diameter, implying each circle has diameter 2 and radius 1; each therefore has area \(\pi \times 1^2 = \pi\). Twenty-four such circles have total area \(24\pi\). The covered fraction is therefore \(24\pi / 36\pi = 2/3\).
ANSWER 3: B
Problem 4:
Within each division the number of games is \(\binom{6}{2} \times 2 = 30\); with two divisions this contributes \(60\) games. Between divisions every team of one division plays every team of the other once, giving \(6 \times 6 = 36\) games. The total is therefore \(60 + 36 = 96\).
ANSWER 4: B
Problem 5:
The daily totals form an arithmetic sequence whose \(n\)th term is the \(n\)th odd positive integer: \(2n-1\). The sum of the first 20 odd positives equals \(20^2 = 400\).
ANSWER 5: D
Problem 6:
A 4.2-megabyte file contains \(4.2 \times 8000 = 33600\) kilobits. At 56 kilobits per second the download time is \(33600/56 = 600\) seconds, or exactly 10 minutes.
ANSWER 6: B
Problem 7:
Exactly one of the three statements is true. The assumption that statement I is the true one forces Amy to be oldest, contradicting I. The assumption that statement II is the true one forces Bill to be oldest, contradicting the requirement that I be false. Hence statement III must be the sole true statement, which forces Amy to be oldest, Celine to be middle, and Bill to be youngest. This ordering satisfies exactly one true statement.
ANSWER 7: E
Problem 8:
The available digits are 0, 1, 2, 2. The admissible 4-digit numbers (first digit nonzero) are obtained by exhaustive enumeration:
starting with 1: 1022, 1202, 1220 (3 numbers);
starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers).
The total is 9.
ANSWER 8: D
Problem 9:
There are \(P(4,3) = 24\) possible three-digit numbers. A number is divisible by 3 precisely when the sum of its digits is divisible by 3. The only admissible digit triples whose sums are multiples of 3 are {1,2,3} and {2,3,4}. Each triple admits 6 permutations, giving 12 favorable numbers and probability \(12/24 = 1/2\).
ANSWER 9: C
Problem 10:
Set Pedro’s finish time to 0. Olympia finishes 3 minutes later, Melina 2 minutes after Olympia, Luke 6 minutes after Olympia, and Nico 11 minutes after Pedro. The ordered finish times are therefore Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Luke finishes fourth.
ANSWER 10: A
Problem 11:
Equating values yields the equation \(20 \times 0.25 + 10 \times 0.10 = 10 \times 0.25 + n \times 0.10\). Simplifying gives \(6 = 2.5 + 0.1n\), so \(n = 35\).
ANSWER 11: D
Problem 12:
After \(m\) months Brent has \(4 \times 4^m = 4^{m+1}\) fish and Gretel has \(128 \times 2^m = 2^{m+7}\) fish. Setting the expressions equal produces the exponent equation \(2m+2 = m+7\), whose solution is \(m = 5\).
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem and understand what is being asked.
We need to find the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in the given addition problem.

## Step 2: Problem 1 - Set up the math to minimize the sum.
To minimize the sum, we should place the smaller numbers in the hundreds places and the larger numbers in the ones places. The two 3-digit numbers can be represented as abc and def, where a, b, c, d, e, and f are the digits 4, 5, 6, 7, 8, 9.

## Step 3: Problem 1 - Assign the digits to minimize the sum.
To get the smallest sum, we should let a = 4 and d = 5 (the two smallest numbers for the hundreds places), b = 6 and e = 7 (the next smallest for the tens places), and c = 8 and f = 9 (the largest for the ones places). So, the numbers are 468 and 579.

## Step 4: Problem 1 - Calculate the sum.
The sum = 468 + 579 = 1047.

## Step 5: Problem 1 - Rule out the tempting wrong choices and confirm the answer.
Let's check the other options by considering the next possible arrangement. If we swap the hundreds digits (4 and 5) with the tens digits (6 and 7), we get 478 + 569 = 1047, which is the same. Other arrangements will yield larger sums. So, 1047 is indeed the smallest sum.

## Step 6: Problem 1 - Provide the final answer.
ANSWER 1: C

## Step 7: Problem 2 - Restate the problem and understand what is being asked.
We need to determine how many half-dozens of eggs Walter should buy to make enough cookies for 108 students who eat an average of 2 cookies apiece, given that the recipe makes 15 cookies and uses 2 eggs per pan.

## Step 8: Problem 2 - Calculate the total number of cookies needed.
Total cookies = 108 students * 2 cookies/student = 216 cookies.

## Step 9: Problem 2 - Determine the number of pans needed.
Number of pans = Total cookies / cookies per pan = 216 / 15 = 14.4. Since only full recipes are made, we need 15 pans.

## Step 10: Problem 2 - Calculate the total number of eggs needed.
Total eggs = 15 pans * 2 eggs/pan = 30 eggs.

## Step 11: Problem 2 - Determine the number of half-dozens of eggs to buy.
Since there are 6 eggs in a half-dozen, the number of half-dozens needed = Total eggs / 6 = 30 / 6 = 5.

## Step 12: Problem 2 - Provide the final answer.
ANSWER 2: C

## Step 13: Problem 3 - Restate the problem and understand what is being asked.
We are given a 12-inch pizza and told that 6 pepperoni circles fit across its diameter. We need to find the fraction of the pizza covered by 24 pepperoni circles.

## Step 14: Problem 3 - Calculate the radius of the pizza and the pepperoni.
The diameter of the pizza is 12 inches, so its radius is 6 inches. The diameter of a pepperoni circle is 12 / 6 = 2 inches, so its radius is 1 inch.

## Step 15: Problem 3 - Calculate the area of the pizza and the total area of the pepperoni circles.
Area of the pizza = π * (6)^2 = 36π. Area of one pepperoni = π * (1)^2 = π. Total area of 24 pepperoni circles = 24 * π = 24π.

## Step 16: Problem 3 - Calculate the fraction of the pizza covered by pepperoni.
Fraction covered = Total area of pepperoni / Area of pizza = 24π / 36π = 2/3.

## Step 17: Problem 3 - Provide the final answer.
ANSWER 3: B

## Step 18: Problem 4 - Restate the problem and understand what is being asked.
The Little Twelve Basketball League has 2 divisions with 6 teams each. Each team plays every other team in its division twice and every team in the other division once. We need to find the total number of games scheduled.

## Step 19: Problem 4 - Calculate the number of games within a division.
In one division, each of the 6 teams plays 5 other teams twice, so the total number of games within a division = (6 * 5) / 2 * 2 = 30. Since there are 2 divisions, the total number of games within divisions = 2 * 30 = 60.

## Step 20: Problem 4 - Calculate the number of games between divisions.
Each team plays 6 teams from the other division once, so for 6 teams, it's 6 * 6 = 36 games. Since there are 6 teams in each division, the total number of games between divisions = 6 * 6 = 36.

## Step 21: Problem 4 - Calculate the total number of games scheduled.
Total games = Games within divisions + Games between divisions = 60 + 36 = 96.

## Step 22: Problem 4 - Provide the final answer.
ANSWER 4: B

## Step 23: Problem 5 - Restate the problem and understand what is being asked.
Janabel sells widgets in a sequence where the number of widgets sold each day increases by 2, starting with 1 on the first day. We need to find the total number of widgets sold after 20 days.

## Step 24: Problem 5 - Identify the sequence of widgets sold.
The sequence is 1, 3, 5, ..., for 20 terms. This is an arithmetic sequence with the first term a = 1 and common difference d = 2.

## Step 25: Problem 5 - Calculate the sum of the sequence.
The nth term = a + (n-1)d = 1 + (n-1)2 = 2n - 1. The sum of the first n terms = n/2 * (a + nth term) = n/2 * (1 + 2n - 1) = n^2. For n = 20, the sum = 20^2 = 400.

## Step 26: Problem 5 - Provide the final answer.
ANSWER 5: D

## Step 27: Problem 6 - Restate the problem and understand what is being asked.
We need to find out how many minutes it would take to download a 4.2-megabyte song at a speed of 56 kilobits per second.

## Step 28: Problem 6 - Convert the size of the song into kilobits.
Since there are 8000 kilobits in a megabyte, the song size in kilobits = 4.2 * 8000 = 33600 kilobits.

## Step 29: Problem 6 - Calculate the download time in seconds.
Download time = Total kilobits / kilobits per second = 33600 / 56 = 600 seconds.

## Step 30: Problem 6 - Convert the download time into minutes.
Download time in minutes = 600 / 60 = 10 minutes.

## Step 31: Problem 6 - Provide the final answer.
ANSWER 6: B

## Step 32: Problem 7 - Restate the problem and understand what is being asked.
We are given three statements about the ages of Amy, Bill, and Celine, and we know exactly one of them is true. We need to rank them

**Problem 1:**  
We need the smallest sum of two 3‑digit numbers formed by using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimise the sum, we place the smallest digits in the hundreds places: 4 and 5 (sum of hundreds = 900). The remaining digits are 6, 7, 8, 9. For the smallest possible sum we assign the next smallest digits to the tens places: 6 and 7 (sum = 130) and the largest to the ones places: 8 and 9 (sum = 17). The total sum is \(900 + 130 + 17 = 1047\). (Other placements, such as trying to avoid a carry from the tens, yield larger sums; 1047 is the minimum.)  
**ANSWER 1: C**

**Problem 2:**  
Total cookies required: \(108 \text{ students} \times 2 = 216\) cookies.  
Each pan makes 15 cookies, so pans needed: \(\lceil 216/15 \rceil = 15\) pans.  
Eggs needed: \(15 \times 2 = 30\) eggs.  
Eggs are bought in half‑dozens (6 eggs per half‑dozen): \(\lceil 30/6 \rceil = 5\).  
**ANSWER 2: C**

**Problem 3:**  
Six pepperoni circles fit exactly across the 12‑inch diameter, so the diameter of one pepperoni is \(12/6 = 2\) inches, radius \(1\) inch.  
Area of one pepperoni \(= \pi \times 1^2 = \pi\) in².  
Total pepperoni area \(= 24\pi\) in².  
Pizza area: radius 6 in, area \(= \pi \times 6^2 = 36\pi\) in².  
Fraction covered \(= \frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}\).  
**ANSWER 3: B**

**Problem 4:**  
Each division has 6 teams.  
Games within one division: each pair plays twice \(\rightarrow 2 \times \binom{6}{2} = 2 \times 15 = 30\). Two divisions give \(60\) games.  
Games between divisions: each team plays every team in the other division once \(\rightarrow 6 \times 6 = 36\) games.  
Total games \(= 60 + 36 = 96\).  
**ANSWER 4: B**

**Problem 5:**  
Widgets sold per day: 1, 3, 5, … – an arithmetic sequence with first term \(a = 1\), common difference \(d = 2\).  
After 20 days, total sold \(S_{20} = \frac{20}{2}\bigl(2\times 1 + (20-1)\times 2\bigr) = 10 \times (2+38) = 10 \times 40 = 400\).  
**ANSWER 5: D**

**Problem 6:**  
Size of song: \(4.2 \text{ MB} \times 8000 \text{ kb/MB} = 33600 \text{ kb}\).  
Download speed \(= 56 \text{ kb/s}\).  
Time in seconds \(= 33600 / 56 = 600\) s.  
Time in minutes \(= 600 / 60 = 10\).  
**ANSWER 6: B**

**Problem 7:**  
Exactly one statement is true.  
Test each case:  
- If I is true (Bill oldest), then II (“Amy not oldest”) must be false → Amy is oldest, contradiction.  
- If II is true (Amy not oldest), I false (Bill not oldest) and III false → “Celine not youngest” false means Celine is youngest. Then neither Amy nor Bill is oldest → Celine must be oldest, contradiction.  
- If III is true (Celine not youngest), I false (Bill not oldest) and II false → “Amy not oldest” false means Amy is oldest. Then Amy oldest, Bill not oldest, Celine not youngest → Bill must be youngest, Celine middle. Order: Amy, Celine, Bill.  
**ANSWER 7: E**

**Problem 8:**  
The digits of 2012 are 2, 0, 1, 2. Total distinct permutations: \(\frac{4!}{2!} = 12\).  
Numbers starting with 0 are not 4‑digit: remaining digits 2,1,2 → \(\frac{3!}{2!} = 3\).  
Valid 4‑digit numbers: \(12 - 3 = 9\).  
**ANSWER 8: D**

**Problem 9:**  
The four digits are 1, 2, 3, 4. A 3‑digit number is a multiple of 3 iff the sum of its digits is divisible by 3.  
Total sum of all four digits = 10. Leaving one digit out, the sum of the chosen three is \(10 - x\).  
We need \(10 - x \equiv 0 \pmod 3 \Rightarrow x \equiv 1 \pmod 3\). Among {1,2,3,4}, \(x = 1\) or \(4\) work.  
There are \(\binom{4}{3}=4\) equally likely sets of three digits, and each set gives equally many permutations. Two sets yield a multiple of 3. Probability \(= \frac{2}{4} = \frac{1}{2}\).  
**ANSWER 9: C**

**Problem 10:**  
Set Pedro’s time = 0 (fastest).  
Nico: 11 minutes behind → Nico’s time = 11.  
Olympia: 3 minutes behind Pedro → Olympia = 3.  
Olympia finished 2 minutes ahead of Melina → Melina = 3 + 2 = 5.  
Olympia finished 6 minutes ahead of Luke → Luke = 3 + 6 = 9.  
Order (fastest to slowest): Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).  
Fourth place = Luke.  
**ANSWER 10: A**

**Problem 11:**  
Value of 20 quarters + 10 dimes: \(20\times 25 + 10\times 10 = 500 + 100 = 600\) cents.  
Value of 10 quarters + \(n\) dimes: \(10\times 25 + 10n = 250 + 10n\).  
Set equal: \(250 + 10n = 600 \Rightarrow 10n = 350 \Rightarrow n = 35\).  
**ANSWER 11: D**

**Problem 12:**  
Brent’s fish: starts at 4, quadruples each month → after \(t\) months: \(B = 4 \cdot 4^t = 4^{t+1} = 2^{2t+2}\).  
Gretel’s fish: starts at 128, doubles each month → after \(t\) months: \(G = 128 \cdot 2^t = 2^7 \cdot 2^t = 2^{t+7}\).  
Set equal: \(2t+2 = t+7 \Rightarrow t = 5\) months.  
**ANSWER 12: B**

Problem 1:
We need to place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers to minimize their sum. Let the numbers be $\overline{ABC}$ and $\overline{DEF}$.

The sum equals $100(A+D) + 10(B+E) + (C+F)$.

To minimize this, we assign the smallest digits to the highest place values:
- **Hundreds place** (weight 100): use 4 and 5 → contribution = $100(4+5) = 900$
- **Tens place** (weight 10): use 6 and 7 → contribution = $10(6+7) = 130$
- **Ones place** (weight 1): use 8 and 9 → contribution = $8+9 = 17$

Minimum sum = $900 + 130 + 17 = 1047$.

Example: $468 + 579 = 1047$.

Ruling out other choices: 947 is impossible since the minimum hundreds contribution alone is 900 and the remaining digits sum to at least $6+7+8+9 = 30$, giving at least 930. Swapping any larger digit into a higher place value increases the sum (e.g., 1056 comes from putting 8 in the tens place).

ANSWER 1: C

Problem 2:
Total cookies needed: $108 \times 2 = 216$ cookies.

Each pan makes 15 cookies. Pans needed: $216 \div 15 = 14.4$. Since only full recipes are made, we need **15 pans**.

Each pan uses 2 eggs: $15 \times 2 = 30$ eggs needed.

Eggs are bought by the half-dozen (6 eggs): $30 \div 6 = 5$ half-dozens.

Ruling out wrong answers: Choice A (1) or B (2) give far too few eggs. Choice E (15) would be the number of pans, not half-dozens of eggs.

ANSWER 2: C

Problem 3:
Six pepperoni circles fit across the 12-inch diameter, so each pepperoni has diameter $12 \div 6 = 2$ inches, giving radius $r = 1$ inch.

Area of one pepperoni circle: $\pi(1)^2 = \pi$ sq inches.
Area of 24 pepperoni circles: $24\pi$ sq inches.

Area of the 12-inch pizza (radius 6): $\pi(6)^2 = 36\pi$ sq inches.

Fraction covered: $\frac{24\pi}{36\pi} = \frac{24}{36} = \frac{2}{3}$.

Ruling out others: $\frac{1}{2}$ would be 18 pepperoni; $\frac{3}{4}$ would be 27 pepperoni.

ANSWER 3: B

Problem 4:
**Within each division:** 6 teams, each pair plays twice.
Number of pairs: $\binom{6}{2} = 15$. Games per division: $15 \times 2 = 30$.
Two divisions: $30 \times 2 = 60$ games.

**Between divisions:** Each of the 6 teams in Division 1 plays each of the 6 teams in Division 2 once.
Inter-division games: $6 \times 6 = 36$.

Total games: $60 + 36 = 96$.

Ruling out others: 80 misses some games; 108 double-counts inter-division games; 192 double-counts everything.

ANSWER 4: B

Problem 5:
Daily sales form an arithmetic sequence: 1, 3, 5, 7, ...
On day $n$, she sells $2n - 1$ widgets.

Total after 20 days = sum of first 20 odd numbers:
$$\sum_{n=1}^{20}(2n-1) = 20^2 = 400.$$

Alternatively, using the arithmetic series formula: $S = \frac{n}{2}(\text{first} + \text{last}) = \frac{20}{2}(1 + 39) = 10 \times 40 = 400$.

Ruling out others: 401 is $20^2 + 1$ (off-by-one error); 210 is $\frac{20 \times 21}{2}$ (sum of 1 to 20, wrong sequence); 39 is the last day's sales only.

ANSWER 5: D

Problem 6:
File size: $4.2$ megabytes $= 4.2 \times 8000 = 33{,}600$ kilobits.

Download speed: 56 kilobits per second.

Time in seconds: $\frac{33{,}600}{56} = 600$ seconds.

Time in minutes: $\frac{600}{60} = 10$ minutes.

Ruling out others: 0.6 would result from forgetting to convert seconds to minutes properly; 1800 and 7200 come from unit conversion errors (e.g., confusing bits and bytes).

ANSWER 6: B

Problem 7:
Exactly one of three statements is true:
- I. Bill is the oldest.
- II. Amy is not the oldest.
- III. Celine is not the youngest.

**Case: Only III is true.**
- I false → Bill is NOT the oldest.
- II false → Amy IS the oldest.
- III true → Celine is NOT the youngest.

Since Amy is oldest, and Celine is not youngest, Celine must be middle. That leaves Bill as youngest.
Order: **Amy, Celine, Bill**.

Verification: I (Bill oldest?) False ✓; II (Amy not oldest?) False ✓; III (Celine not youngest?) True ✓. Exactly one true.

Testing other cases leads to contradictions (e.g., if only I is true, then II false means Amy is also oldest — impossible with different ages).

ANSWER 7: E

Problem 8:
The digits of 2012 are: 2, 0, 1, 2 (two 2's, one 0, one 1).

Total 4-digit arrangements: $\frac{4!}{2!} = 12$.

Numbers starting with 0 are not valid 4-digit numbers (they're 3-digit numbers ≤ 999, hence not > 1000).
Arrangements starting with 0: arrange {2, 1, 2} in 3 spots = $\frac{3!}{2!} = 3$.

Valid numbers: $12 - 3 = 9$.

Listing them confirms:
- Starting with 1: 1022, 1202, 1220 (3 numbers)
- Starting with 2: 2012, 2021, 2102, 2120, 2201, 2210 (6 numbers)
Total = 9.

ANSWER 8: D

Problem 9:
We draw 3 digits from {1, 2, 3, 4}. A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The $\binom{4}{3} = 4$ possible digit sets:
- {1, 2, 3}: sum = 6 → divisible by 3 ✓
- {1, 2, 4}: sum = 7 → not divisible by 3 ✗
- {1, 3, 4}: sum = 8 → not divisible by 3 ✗
- {2, 3, 4}: sum = 9 → divisible by 3 ✓

Each set is equally likely to be drawn. 2 out of 4 sets give multiples of 3.

Probability = $\frac{2}{4} = \frac{1}{2}$.

Note: The order of drawing doesn't affect divisibility by 3, so we only need to consider which subset is chosen.

ANSWER 9: C

Problem 10:
Let Pedro's finish time be $P$. Express all times relative to $P$:
- Pedro: $P$
- Nico: $P + 11$ (11 min behind Pedro)
- Olympia: $P + 3$ (3 min behind Pedro)
- Melina: Olympia + 2 = $P + 5$ (Olympia was 2 min ahead of Melina)
- Luke: Olympia + 6 = $P + 9$ (Olympia was 6 min ahead of Luke)

Ranking by finish time (smallest = first):
1. Pedro ($P$)
2. Olympia ($P + 3$)
3. Melina ($P + 5$)
4. **Luke** ($P + 9$)
5. Nico ($P + 11$)

Fourth place is Luke.

ANSWER 10: A

Problem 11:
Value of 20 quarters and 10 dimes:
$20(25) + 10(10) = 500 + 100 = 600$ cents.

Value of 10 quarters and $n$ dimes:
$10(25) + n(10) = 250 + 10n$ cents.

Setting equal:
$250 + 10n = 600$
$10n = 350$
$n = 35$.

Ruling out others: 10 or 20 would come from incorrectly assuming a 1-to-1 swap; 45 would result from adding instead of subtracting.

ANSWER 11: D

Problem 12:
Brent starts with 4 goldfish, quadruples monthly. After $m$ months: 

 **Problem 1:**
We must place the digits 4, 5, 6, 7, 8, 9 into two 3-digit numbers so their sum is as small as possible.  
If the numbers are \(ABC\) and \(DEF\), their sum is  
\[100(A+D)+10(B+E)+(C+F).\]  
To minimize the total, assign the smallest available digits to the highest place values.  
- Hundreds places: use 4 and 5, contributing \(100(4+5)=900\).  
- Tens places: use 6 and 7, contributing \(10(6+7)=130\).  
- Ones places: use 8 and 9, contributing \(8+9=17\).  

The minimum sum is \(900+130+17=1047\).  
(Any other pairing gives a larger hundreds contribution and thus a larger total.)

ANSWER 1: C

---

**Problem 2:**
First find the total cookies needed:  
\(108\text{ students}\times 2\text{ cookies/student}=216\text{ cookies}\).  

Each full recipe makes 15 cookies. Since only full recipes are allowed:  
\(\lceil 216/15\rceil = \lceil 14.4\rceil = 15\) recipes.  

Each recipe uses 2 eggs, so eggs needed:  
\(15\times 2 = 30\) eggs.  

Eggs are bought by the half-dozen (6 eggs):  
\(30\div 6 = 5\) half-dozens exactly.

ANSWER 2: C

---

**Problem 3:**
The pizza has diameter 12 inches, so radius 6 inches. Its area is \(\pi(6^2)=36\pi\).  
Six pepperoni circles fit exactly across the diameter, so each pepperoni has diameter \(12/6=2\) inches and radius 1 inch.  
Area of one pepperoni: \(\pi(1^2)=\pi\).  

With 24 non-overlapping pepperoni circles, total pepperoni area is \(24\pi\).  
Fraction of pizza covered:  
\[\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.\]

ANSWER 3: B

---

**Problem 4:**
There are two divisions of 6 teams each.

*Within one division:* Each team plays the other 5 teams twice.  
Number of games = \(\binom{6}{2}\times 2 = 15\times 2 = 30\).  
For two divisions: \(30+30=60\) games.

*Between divisions:* Each of the 6 teams in one division plays each of the 6 teams in the other division once.  
Number of games = \(6\times 6 = 36\).

Total scheduled games: \(60+36=96\).

ANSWER 4: B

---

**Problem 5:**
Janabel’s daily sales form an arithmetic sequence: \(1, 3, 5, \dots\)  
First term \(a_1=1\), common difference \(d=2\), number of days \(n=20\).  

The 20th term is  
\[a_{20}=1+(20-1)\cdot 2 = 39.\]  

Total widgets sold:  
\[S_{20}=\frac{20}{2}(1+39)=10\cdot 40=400.\]

ANSWER 5: D

---

**Problem 6:**
Convert the song size to kilobits:  
\(4.2\text{ MB}\times 8000\text{ kb/MB}=33\,600\text{ kilobits}\).  

Download speed is 56 kilobits per second, so time in seconds is  
\[\frac{33\,600}{56}=600\text{ seconds}.\]  

Convert to minutes:  
\(600\div 60 = 10\) minutes.

ANSWER 6: B

---

**Problem 7:**
Exactly one of the three statements is true.  
- I. Bill is the oldest.  
- II. Amy is not the oldest.  
- III. Celine is not the youngest.  

Test the rankings:

*If Bill were oldest (I true), then II (“Amy is not oldest”) would also be true. Two true statements—impossible.*  
Thus I is false, so Bill is **not** the oldest.

Now exactly one of II or III must be true.

*Suppose II is true (Amy is not oldest) and III is false (Celine is youngest).*  
Since Bill is not oldest and Amy is not oldest, Celine would have to be oldest—but III false says she is youngest. Contradiction.

*Therefore II is false and III is true.*  
II false means Amy **is** the oldest.  
III true means Celine is not the youngest, so Celine is middle and Bill is youngest.  

Ranking: Amy, Celine, Bill.

ANSWER 7: E

---

**Problem 8:**
The digits available are 2, 0, 1, 2 (two 2’s). We need 4-digit numbers > 1000, so the thousands digit cannot be 0.

**Thousands digit = 1:**  
Remaining digits to arrange: 2, 0, 2.  
Distinct permutations: \(\frac{3!}{2!}=3\) numbers (1022, 1202, 1220).

**Thousands digit = 2:**  
Remaining digits to arrange: 0, 1, 2.  
Distinct permutations: \(3!=6\) numbers (2012, 2021, 2102, 2120, 2201, 2210).

Total: \(3+6=9\).

ANSWER 8: D

---

**Problem 9:**
Total possible three-digit numbers: choose an ordered sequence of 3 digits from {1,2,3,4} without replacement.  
Count = \(4\cdot 3\cdot 2 = 24\).

A number is divisible by 3 exactly when the sum of its digits is divisible by 3.  
Possible 3-element subsets:
- {1,2,3}: sum 6 (good)
- {1,2,4}: sum 7 (bad)
- {1,3,4}: sum 8 (bad)
- {2,3,4}: sum 9 (good)

Each good set can be arranged in \(3!=6\) ways.  
Favorable outcomes: \(6+6=12\).

Probability: \(\frac{12}{24}=\frac12\).

ANSWER 9: C

---

**Problem 10:**
Let Pedro’s time be \(P\) minutes (smaller is faster/better).

- Nico: \(P+11\)
- Olympia: \(P+3\)
- Melina: \((P+3)+2 = P+5\) (2 minutes behind Olympia)
- Luke: \((P+3)+6 = P+9\) (6 minutes behind Olympia)

Ordering from first to last:
1. Pedro (\(P\))
2. Olympia (\(P+3\))
3. Melina (\(P+5\))
4. Luke (\(P+9\))
5. Nico (\(P+11\))

The fourth-place runner is Luke.

ANSWER 10: A

---

**Problem 11:**
Value in cents:
- 20 quarters = \(20\times 25 = 500\) cents
- 10 dimes = \(10\times 10 = 100\) cents
- Left side total = 600 cents

Right side:
- 10 quarters = \(10\times 25 = 250\) cents
- \(n\) dimes = \(10n\) cents

Equation:  
\(600 = 250 + 10n\)  
\(350 = 10n\)  
\(n = 35\).

ANSWER 11: D

---

**Problem 12:**
Let \(m\) be the number of months from now.

- Brent: starts with 4, quadruples monthly \(\rightarrow 4\cdot 4^{m}=4^{m+1}\)
- Gretel: starts with 128, doubles monthly \(\rightarrow 128\cdot 2^{m}\)

Set them equal:  
\(4\cdot 4^{m} = 128\cdot 2^{m}\)  
Write as powers of 2:  
\(2^{2}\cdot (2^{2})^{m} = 2^{7}\cdot 2^{m}\)  
\(2^{2m+2} = 2^{m+7}\)

Equate exponents:  
\(2m+2 = m+7\)  
\(m = 5\).

Check: Brent \(4\cdot 4^{5}=4096\); Gretel \(128\cdot 2^{5}=4096\).

ANSWER 12: B

Problem 1:
To find the smallest sum of two 3-digit numbers using digits 4, 5, 6, 7, 8, 9, I need to minimize each place value from highest to lowest, while accounting for carries.

**Hundreds place:** Use the two smallest digits (4 and 5) → sum = 9
**Tens place:** Use the next two smallest (6 and 7) → sum = 13
**Ones place:** Use the remaining two (8 and 9) → sum = 17

Now computing with carries:
- Ones: 8 + 9 = 17, write 7, carry 1
- Tens: 6 + 7 + 1 = 14, write 4, carry 1
- Hundreds: 4 + 5 + 1 = 10, write 10

Sum = 1047

Verification: 468 + 579 = 1047 ✓

Any rearrangement within each place value gives the same sum, and using larger digits in higher places would increase the sum.

ANSWER 1: C

---

Problem 2:
**Total cookies needed:** 108 students × 2 cookies = 216 cookies

**Number of full recipes (pans):** ⌈216/15⌉ = ⌈14.4⌉ = 15 pans

**Total eggs needed:** 15 pans × 2 eggs/pan = 30 eggs

**Half-dozens of eggs:** 30 ÷ 6 = 5 half-dozens

ANSWER 2: C

---

Problem 3:
**Pepperoni diameter:** 6 circles fit across 12-inch diameter → each pepperoni has diameter = 12/6 = 2 inches, radius = 1 inch

**Area of one pepperoni:** π(1)² = π

**Total pepperoni area:** 24 × π = 24π

**Pizza radius:** 6 inches, area = π(6)² = 36π

**Fraction covered:** 24π/36π = 2/3

ANSWER 3: B

---

Problem 4:
**Within each division:** Each team plays the other 5 teams twice.
Games per division = C(6,2) × 2 = 15 × 2 = 30
Games in both divisions = 30 × 2 = 60

**Between divisions:** Each of 6 teams in Division 1 plays each of 6 teams in Division 2 once.
Games between divisions = 6 × 6 = 36

**Total games:** 60 + 36 = 96

ANSWER 4: B

---

Problem 5:
Janabel's daily sales: 1, 3, 5, 7, ..., (2n-1) for day n.

After 20 days, she sells: 1 + 3 + 5 + ... + 39

The sum of the first n odd numbers equals n².

Total = 20² = 400

ANSWER 5: D

---

Problem 6:
**Song size in kilobits:** 4.2 megabytes × 8000 kilobits/megabyte = 33,600 kilobits

**Download time in seconds:** 33,600 ÷ 56 = 600 seconds

**Convert to minutes:** 600 ÷ 60 = 10 minutes

ANSWER 6: B

---

Problem 7:
Testing which single statement being true is consistent:

**If I is true (Bill is oldest):** Then II must be false → Amy IS the oldest. Contradiction with Bill being oldest.

**If II is true (Amy is not oldest):** Then III must be false → Celine IS the youngest. And I must be false → Bill is NOT the oldest. So Celine is oldest (since Amy isn't and Bill isn't). But Celine can't be both oldest and youngest. Contradiction.

**If III is true (Celine is not youngest):** Then I must be false → Bill is NOT the oldest. And II must be false → Amy IS the oldest. So Amy is oldest, Celine is not youngest (so Bill is youngest), and Celine is in the middle.

Order: Amy (oldest), Celine, Bill (youngest)

Verification: I is false ✓, II is false ✓, III is true ✓ — exactly one true statement.

ANSWER 7: E

---

Problem 8:
Digits of 2012: 0, 1, 2, 2

**Numbers starting with 1:** Permutations of {0, 2, 2} = 3!/2! = 3
(1022, 1202, 1220)

**Numbers starting with 2:** Permutations of {0, 1, 2} = 3! = 6
(2012, 2021, 2102, 2120, 2201, 2210)

All are greater than 1000.

**Total:** 3 + 6 = 9

ANSWER 8: D

---

Problem 9:
Total 3-digit numbers possible: 4 × 3 × 2 = 24

A number is divisible by 3 if the sum of its digits is divisible by 3.

**Possible digit combinations and their sums:**
- {1,2,3}: sum = 6 ✓
- {1,2,4}: sum = 7 ✗
- {1,3,4}: sum = 8 ✗
- {2,3,4}: sum = 9 ✓

**Favorable outcomes:** Each valid set gives 3! = 6 arrangements, so 2 × 6 = 12

**Probability:** 12/24 = 1/2

ANSWER 9: C

---

Problem 10:
Let P = Pedro's finish time. Lower time = faster finish.

- Nico: P + 11 (11 minutes behind Pedro)
- Olympia: P - 3 (3 minutes behind Pedro)
- Melina: O + 2 = P - 1 (2 minutes behind Olympia)
- Luke: O + 6 = P + 3 (6 minutes behind Olympia)

**Ranking (fastest to slowest):**
1. Olympia: P - 3
2. Melina: P - 1
3. Pedro: P
4. Luke: P + 3
5. Nico: P + 11

Fourth place is Luke.

ANSWER 10: A

---

Problem 11:
Setting the values equal (quarter = 25¢, dime = 10¢):

20(25) + 10(10) = 10(25) + n(10)
500 + 100 = 250 + 10n
600 = 250 + 10n
350 = 10n
n = 35

ANSWER 11: D

---

Problem 12:
After t months:
- Brent: 4 × 4^t
- Gretel: 128 × 2^t

Setting equal:
4 × 4^t = 128 × 2^t
4 × (2²)^t = 128 × 2^t
4 × 2^(2t) = 128 × 2^t
2² × 2^(2t) = 2^7 × 2^t
2^(2t+2) = 2^(t+7)
2t + 2 = t + 7
t = 5

ANSWER 12: B

Problem 1:
We have six digits 4, 5, 6, 7, 8, 9 to place in the six boxes of a 3‑digit + 3‑digit addition.  
The sum of the two numbers equals  

\[
100(A+D)+10(B+E)+(C+F)
\]

where \((A,B,C)\) and \((D,E,F)\) are the digits of the two numbers.  
To minimize the sum we want the smallest digits in the hundreds places (weight 100), the next smallest in the tens (weight 10) and the largest in the ones (weight 1).  

Sorting the digits: 4 < 5 < 6 < 7 < 8 < 9.  
Pair the smallest two for the hundreds column, the next two for the tens column, and the two largest for the ones column:

\[
\begin{aligned}
\text{hundreds:}&\;4+5=9,\\
\text{tens:}&\;6+7=13,\\
\text{ones:}&\;8+9=17.
\end{aligned}
\]

Hence the minimal sum is  

\[
100\cdot9+10\cdot13+17=900+130+17=1047.
\]

No other pairing can give a smaller weighted sum because any other arrangement puts a larger digit in a higher‑weight column, increasing the total.  

ANSWER 1: C

---

Problem 2:
Each student eats 2 cookies, so the school needs  

\[
108\times2=216\text{ cookies}.
\]

A recipe makes 15 cookies per pan and uses 2 eggs per pan.  
The number of full pans required is  

\[
\lceil 216/15\rceil = \lceil 14.4\rceil = 15\text{ pans}.
\]

Eggs needed: \(15\text{ pans}\times2\text{ eggs/pan}=30\) eggs.  
Walter buys eggs by the half‑dozen (6 eggs each).  

\[
30\text{ eggs}/6 = 5\text{ half‑dozens}.
\]

Thus he must buy 5 half‑dozens.  

ANSWER 2: C

---

Problem 3:
Six pepperoni circles fit across the 12‑inch diameter, so each circle’s diameter is  

\[
\frac{12\text{ in}}{6}=2\text{ in}\;\Rightarrow\;r=1\text{ in}.
\]

Area of one pepperoni circle: \(\pi r^{2}=\pi\).  
Total area of 24 circles: \(24\pi\).  

Area of the pizza (radius 6 in): \(\pi(6)^{2}=36\pi\).  

Fraction covered by pepperoni:

\[
\frac{24\pi}{36\pi}=\frac{24}{36}=\frac{2}{3}.
\]

ANSWER 3: B

---

Problem 4:
Intradivision games: each division has \(\binom{6}{2}=15\) pairs of teams, and each pair plays twice, giving  

\[
15\times2=30\text{ games per division}.
\]

Two divisions: \(30\times2=60\) intradivision games.

Interdivision games: each of the 6 teams in one division plays each of the 6 teams in the other division once, giving  

\[
6\times6=36\text{ interdivision games}.
\]

Total games scheduled: \(60+36=96\).  

ANSWER 4: B

---

Problem 5:
Janabel sells \(1,3,5,\dots\) widgets – the odd numbers.  
The \(n\)th day she sells \(2n-1\) widgets.  

Total after 20 days:

\[
\sum_{n=1}^{20}(2n-1)=2\sum_{n=1}^{20}n-\sum_{n=1}^{20}1
=2\cdot\frac{20\cdot21}{2}-20=420-20=400.
\]

(The sum of the first 20 odd numbers is \(20^{2}=400\).)  

ANSWER 5: D

---

Problem 6:
Download speed \(=56\) kilobits / second.  
Song size \(=4.2\) megabytes.  
Since \(1\) megabyte = 8000 kilobits,

\[
4.2\text{ MB}=4.2\times8000=33\,600\text{ kilobits}.
\]

Time in seconds:

\[
\frac{33\,600}{56}=600\text{ s}.
\]

Convert to minutes: \(600\text{ s}/60=10\) minutes.  

ANSWER 6: B

---

Problem 7:
Let the ages be ordered from oldest to youngest.  
Exactly one of the statements I, II, III is true.

- I true ⇒ Bill is oldest.  
- II true ⇒ Amy is not oldest.  
- III true ⇒ Celine is not youngest.

Test each possible ranking (6 permutations). Only the order **Amy, Celine, Bill** makes exactly one statement true (statement III).  

Thus the ranking from oldest to youngest is Amy, Celine, Bill.  

ANSWER 7: E

---

Problem 8:
Digits of 2012 are \(\{2,0,1,2\}\).  
We need 4‑digit numbers > 1000, i.e., the leading digit cannot be 0.  

Total distinct permutations of the four digits: \(4!/2!=12\).  
Those with leading 0 are three (0 + remaining three digits in any order).  

Thus valid numbers: \(12-3=9\).  

ANSWER 8: D

---

Problem 9:
We draw three distinct digits from \(\{1,2,3,4\}\) to form a 3‑digit number.  
Total possible numbers: \(P(4,3)=4\cdot3\cdot2=24\).  

A number is a multiple of 3 iff the sum of its digits is a multiple of 3.  

The sum of the chosen three digits is a multiple of 3 exactly when the omitted digit is 1 or 4 (the digits congruent to 1 (mod 3)).  
Thus the favorable subsets are \(\{2,3,4\}\) and \(\{1,2,3\}\).  

Each subset yields \(3!=6\) permutations, so favorable outcomes \(=2\times6=12\).  

Probability \(=12/24=1/2\).  

ANSWER 9: C

---

Problem 10:
Let the finishing times be \(T_{\text{L}}, T_{\text{M}}, T_{\text{N}}, T_{\text{O}}, T_{\text{P}}\).  
Given:

\[
\begin{aligned}
T_{\text{N}} &= T_{\text{P}}+11,\\
T_{\text{O}} &= T_{\text{P}}+3,\\
T_{\text{O}} &= T_{\text{M}}-2 \;\Rightarrow\; T_{\text{M}}=T_{\text{P}}+5,\\
T_{\text{O}} &= T_{\text{L}}-6 \;\Rightarrow\; T_{\text{L}}=T_{\text{P}}+9.
\end{aligned}
\]

Thus (from earliest to latest):

\[
T_{\text{P}}=0,\; T_{\text{O}}=3,\; T_{\text{M}}=5,\; T_{\text{L}}=9,\; T_{\text{N}}=11.
\]

The fourth place runner is Luke.  

ANSWER 10: A

---

Problem 11:
Quarters are worth 25¢, dimes 10¢.  

\[
20\text{ quarters}+10\text{ dimes}=20(25)+10(10)=500+100=600\text{¢}.
\]

Set equal to \(10\) quarters + \(n\) dimes:

\[
600=10(25)+n(10)\;\quad 600=250+10n\;\Rightarrow\;10n=350\;\Rightarrow\;n=35.
\]

ANSWER 11: D

---

Problem 12:
Brent’s goldfish quadruple each month: after \(t\) months he has  

\[
4\cdot4^{t}=4^{\,t+1}.
\]

Gretel’s goldfish double each month: after \(t\) months she has  

\[
128\cdot2^{t}=2^{\,7}\cdot2^{t}=2^{\,t+7}.
\]

Set them equal:

\[
4^{\,t+1}=2^{\,t+7}\;\;(2^{2})^{\,t+1}=2^{\,t+7}\;\;
2^{\,2t+2}=2^{\,t+7}\;\Rightarrow\;2t+2=t+7\;\Rightarrow\;t=5.
\]

Thus after 5 months they have the same number.  

ANSWER 12: B

Problem 1:  
The problem requires finding the smallest sum of two 3-digit numbers using the digits 4, 5, 6, 7, 8, 9 exactly once. To minimize the sum, the smaller digits should occupy the higher place values (hundreds place). Assign the two smallest digits (4 and 5) to the hundreds places. The next smallest digits (6 and 7) go to the tens places, and the largest digits (8 and 9) to the ones places. This gives 468 and 579, but swapping digits to minimize further yields 476 + 589 = 1065 (not optimal). Testing other combinations, the smallest sum is 567 + 489 = 1056, but further optimization gives 468 + 579 = 1047. The correct minimal sum is 1047.  
ANSWER 1: C  

Problem 2:  
108 students eat 2 cookies each, requiring 216 cookies. Each recipe makes 15 cookies and uses 2 eggs. The number of recipes needed is 216 ÷ 15 = 14.4, rounded up to 15 recipes (only full recipes allowed). Total eggs needed: 15 × 2 = 30 eggs. Eggs are sold in half-dozens (6 eggs), so 30 ÷ 6 = 5 half-dozens.  
ANSWER 2: E (Note: The original solution had a calculation error; 15 recipes require 30 eggs, which is 5 half-dozens. However, rechecking: 108 × 2 = 216 cookies; 216 ÷ 15 = 14.4 → 15 pans; 15 × 2 = 30 eggs; 30 ÷ 6 = 5. The answer is E.)  

Problem 3:  
Six pepperoni circles fit across the pizza’s diameter (12 inches), so each pepperoni has a diameter of 2 inches (radius 1 inch). The area of one pepperoni is π(1)² = π. Total pepperoni area: 24π. Pizza area: π(6)² = 36π. Fraction covered: 24π / 36π = 2/3.  
ANSWER 3: B  

Problem 4:  
Each team plays 5 others in its division twice (5 × 2 = 10 games) and 6 teams in the other division once (6 games), totaling 16 games per team. Total games for 12 teams: 12 × 16 = 192, but each game is counted twice, so 192 ÷ 2 = 96 games.  
ANSWER 4: B  

Problem 5:  
Janabel sells widgets in an arithmetic sequence: 1, 3, 5, ..., with a common difference of 2. After 20 days, the total is the sum of the first 20 odd numbers: 20² = 400.  
ANSWER 5: D  

Problem 6:  
Download speed: 56 kilobits/second. Song size: 4.2 megabytes = 4.2 × 8000 = 33,600 kilobits. Time in seconds: 33,600 ÷ 56 = 600 seconds. Convert to minutes: 600 ÷ 60 = 10 minutes.  
ANSWER 6: B  

Problem 7:  
Only one statement is true. Testing options: If Bill is oldest (I true), then II (Amy not oldest) is also true, violating the condition. If III (Celine not youngest) is true, then I and II are false. This implies Amy is oldest, Bill is not oldest, and Celine is not youngest. Order: Amy (oldest), Celine, Bill (youngest).  
ANSWER 7: E  

Problem 8:  
The digits of 2012 are 2, 0, 1, 2. Valid 4-digit numbers >1000: 1022, 1202, 1220, 2012, 2021, 2102, 2120, 2201, 2210. Total: 9.  
ANSWER 8: D  

Problem 9:  
Total 3-digit numbers: 4 × 3 × 2 = 24. Multiples of 3 have digit sums divisible by 3. Valid combinations: 1,2,3 (sum 6); 2,3,4 (sum 9); 1,3,2 (sum 6, already counted); etc. Favorable outcomes: 12. Probability: 12/24 = 1/2.  
ANSWER 9: C (Note: Re-evaluating, the favorable digit sets are {1,2,3} and {2,3,4}, each with 6 permutations, totaling 12. Probability: 12/24 = 1/2.)  

Problem 10:  
Assign variables: Pedro = P, Nico = P + 11, Olympia = P - 3, Melina = P - 5, Luke = P - 9. Order: P, P+11, P-3, P-5, P-9 → Pedro, Nico, Olympia, Melina, Luke. Fourth is Melina.  
ANSWER 10: B  

Problem 11:  
20 quarters + 10 dimes = 10 quarters + n dimes. Value: 20 × 25 + 10 × 10 = 10 × 25 + n × 10. 500 + 100 = 250 + 10n → 600 = 250 + 10n → 10n = 350 → n = 35.  
ANSWER 11: D  

Problem 12:  
Brent’s goldfish: 4 × 4^t. Gretel’s goldfish: 128 × 2^t. Set equal: 4 × 4^t = 128 × 2^t → 4^(t+1) = 2^7 × 2^t → 2^(2t+2) = 2^(t+7) → 2t + 2 = t + 7 → t = 5.  
ANSWER 12: B  

The final answers are:  
ANSWER 1: C  
ANSWER 2: E  
ANSWER 3: B  
ANSWER 4: B  
ANSWER 5: D  
ANSWER 6: B  
ANSWER 7: E  
ANSWER 8: D  
ANSWER 9: C  
ANSWER 10: B  
ANSWER 11: D  
ANSWER 12: B

### Problem 1:
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4,5,6,7,8,9 once to form two 3-digit numbers?
To minimize the sum, we assign the smallest digits to the highest place values (since place values scale in weight):
1. Hundreds places (highest weight) get the two smallest digits: 4 and 5, contributing 900 to the sum.
2. Tens places get the next two smallest digits: 6 and7, contributing 130 to the sum.
3. Ones places get the remaining largest digits:8 and9, contributing 17 to the sum.
Total sum = 900+130+17=1047. Tempting wrong answers: 1037 is unachievable (the smallest possible tens sum is 13, so the tens digit of the total cannot be 3), 947 is too small for two 3-digit numbers.
ANSWER 1: C

---

### Problem 2:
108 students eat 2 cookies each. Full pans of 15 cookies are made, each pan uses 2 eggs. Eggs are sold in half-dozens (6 eggs). How many half-dozens are needed?
Step 1: Total cookies needed = 108*2=216.
Step 2: Number of full pans: ceiling(216/15)=15 pans (14 pans only make 210 cookies, which is insufficient).
Step3: Total eggs needed =15*2=30 eggs.
Step4: Number of half-dozens =30/6=5.
ANSWER 2: C

---

### Problem3:
Six pepperoni fit across the 12-inch diameter of a pizza. 24 pepperoni are placed on the pizza. What fraction of the pizza is covered by pepperoni?
Step1: Pizza radius =6 inches, area=π*6²=36π.
Step2: Each pepperoni diameter=12/6=2 inches, radius=1 inch, area per pepperoni=π*1²=π.
Step3: Total pepperoni area=24π. Fraction=24π/36π=2/3.
ANSWER 3: B

---

### Problem4:
Two divisions of 6 teams each. Teams play every other team in their division twice, and every team in the other division once. Total number of games?
Step1: Intra-division games: For one division, unique pairs of teams = C(6,2)=15. Each pair plays twice: 15*2=30 per division, 60 total for two divisions.
Step2: Inter-division games: 6 teams *6 teams=36 unique cross-division games.
Step3: Total games=60+36=96.
ANSWER 4: B

---

### Problem5:
Janabel sells 1 widget on day1, 3 on day2, 5 on day3, adding 2 widgets per day. Total widgets after 20 days?
This is the sum of the first 20 odd integers, an arithmetic sequence. Sum = n/2*(first term + last term). First term=1, 20th term=1+(19*2)=39. Sum=20/2*(1+39)=10*40=400.
ANSWER 5: D

---

### Problem6:
Download speed 56 kilobits per second. A 4.2-megabyte song, 8000 kilobits=1 megabyte. How many minutes to download?
Step1: Total kilobits of the song=4.2*8000=33600.
Step2: Time in seconds=33600/56=600 seconds.
Step3: Convert to minutes=600/60=10 minutes.
ANSWER 6: B

---

### Problem7:
Exactly one of three statements is true: I.Bill is oldest. II.Amy is not oldest. III.Celine is not youngest. Rank oldest to youngest.
Test which order has exactly one true statement:
- Only if Amy is oldest, Bill is youngest, Celine is middle (order: Amy, Celine, Bill): I is false, II is false, only III is true. This is the only valid case.
ANSWER 7: E

---

### Problem8:
How many 4-digit numbers greater than 1000 use the digits of 2012 (two 2s, one 0, one 1)?
First digit cannot be 0:
- First digit=1: remaining digits 2,0,2 form 3 distinct numbers:1022,1202,1220.
- First digit=2: remaining digits 0,1,2 form 6 distinct permutations:2012,2021,2102,2120,2201,2210.
Total=3+6=9.
ANSWER 8: D

---

### Problem9:
Digits 1,2,3,4. Draw 3 to make a 3-digit number. Probability it is a multiple of 3?
A number is a multiple of 3 if its digit sum is a multiple of 3. All possible triplets of digits: {1,2,3}(sum=6), {1,2,4}(sum=7), {1,3,4}(sum=8), {2,3,4}(sum=9). Only 2 of 4 triplets work, so half of all possible numbers are multiples of 3. Probability=1/2.
ANSWER 9: C

---

### Problem10:
Five runners: Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina, 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Who finished fourth?
Assign Pedro's time as 0:
- O=3 (3 minutes behind P), M=5 (2 minutes behind O), L=9 (6 minutes behind O), N=11 (11 minutes behind P).
Order of finish: 1.Pedro, 2.Olympia,3.Melina,4.Luke,5.Nico. Fourth is Luke.
ANSWER 10: A

---

### Problem11:
20 quarters +10 dimes =10 quarters +n dimes. Find n.
Total value first set:20*25 +10*10=600c. Second set:10*25 +10n=600 → 10n=350 →n=35.
ANSWER 11: D

---

### Problem12:
Brent has 4 goldfish, quadruples monthly. Gretel has 128 goldfish, doubles monthly. When will they have the same number?
Set 4*4ᵗ =128*2ᵗ → 2^(2t+2)=2^(t+7) → 2t+2=t+7 →t=5.
ANSWER 12: B

### Problem 1:
We need the smallest sum of two 3-digit numbers using the digits 4,5,6,7,8,9 exactly once. To minimize the sum, we assign the smallest digits to the highest place values (hundreds, then tens, then ones), since higher place values contribute more to the total sum.
1.  The two smallest digits (4 and 5) are the hundreds digits of the two numbers, so the hundreds place contributes \(100*(4+5) = 900\).
2.  The next two smallest digits (6 and7) are the tens digits, contributing \(10*(6+7)=130\).
3.  The remaining largest digits (8 and9) are the ones digits, contributing \(8+9=17\).
Adding these: \(900 + 130 +17 = 1047\). This is the smallest possible sum, as any other assignment would put larger digits in higher places, increasing the total.
ANSWER 1: C
---
### Problem 2:
1.  Total cookies needed: \(108 \text{ students} * 2 \text{ cookies/student} = 216\) cookies.
2.  Each pan makes 15 cookies, so number of full pans required: \(\lceil 216 /15 \rceil = 15\) (since 14 pans only make 210 cookies, which is insufficient).
3.  Each pan uses 2 eggs, so total eggs needed: \(15 *2 =30\) eggs.
4.  Eggs are bought by the half-dozen (6 eggs per half-dozen), so number of half-dozens: \(30 /6 =5\).
ANSWER 2: C
---
### Problem 3:
1.  The pizza has a 12-inch diameter, so 6 pepperonis fit across the diameter means each pepperoni has diameter \(12/6=2\) inches, so radius 1 inch.
2.  Area of the pizza: \(\pi*(12/2)^2 = 36\pi\) square inches.
3.  Area of one pepperoni: \(\pi*(1)^2 = \pi\) square inches. Total area of 24 pepperonis: \(24*\pi =24\pi\).
4.  Fraction covered: \(24\pi / 36\pi = 24/36 = 2/3\).
ANSWER 3: B
---
### Problem 4:
We calculate intra-division and inter-division games separately:
1.  **Intra-division games**: Each division has 6 teams. The number of unique pairs in a division is \(\binom{6}{2} = 15\). Each pair plays twice, so per division: \(15*2=30\) games. For two divisions: \(30*2=60\).
2.  **Inter-division games**: Each team in one division plays every team in the other division once. There are \(6*6=36\) such pairs, so 36 games.
3.  Total games: \(60 +36=96\).
ANSWER 4: B
---
### Problem 5:
Janabel's daily sales form an arithmetic sequence: first term \(a_1=1\), common difference \(d=2\). The total after 20 days is the sum of the first 20 terms.
1.  The 20th term: \(a_{20} = a_1 + (20-1)d =1 +19*2=39\).
2.  Sum of an arithmetic sequence: \(S_n = \frac{n}{2}*(a_1 +a_n) = \frac{20}{2}*(1+39) =10*40=400\).
(Alternatively, the sum of the first \(k\) odd numbers is \(k^2\), so \(20^2=400\).)
ANSWER 5: D
---
### Problem 6:
1.  Convert the song size to kilobits: \(4.2 \text{ MB} * 8000 \text{ kb/MB} = 33600\) kilobits.
2.  Time to download in seconds: \(\text{total kilobits} / \text{speed} = 33600 /56 = 600\) seconds.
3.  Convert seconds to minutes: \(600 /60 =10\) minutes.
ANSWER 6: B
---
### Problem 7:
We test each case for who is oldest, since exactly one statement is true:
1.  **Case 1: Bill is oldest (I is true)**. Then statement II ("Amy is not the oldest") is also true, which violates the "exactly one true" rule. So this case is impossible.
2.  **Case 2: II is true (Amy is not oldest), so I and III are false**. I false means Bill is not oldest. III false means Celine *is* the youngest. Since Amy is not oldest and Bill is not oldest, Celine must be oldest—but this contradicts Celine being youngest. So this case is impossible.
3.  **Case 3: III is true (Celine is not youngest), so I and II are false**. I false means Bill is not oldest. II false means "Amy is not oldest" is false, so Amy *is* the oldest. With Amy oldest and Celine not youngest, Bill must be the youngest. The order is Amy (oldest), Celine, Bill (youngest).
Check statements: I (Bill oldest) = false, II (Amy not oldest) = false, III (Celine not youngest) = true. Exactly one true, which fits.
ANSWER 7: E
---
### Problem 8:
We count valid 4-digit numbers using digits 2,0,1,2 (two 2s, one 0, one 1), with first digit ≠ 0:
1.  Total permutations of the 4 digits: \(\frac{4!}{2!} = 12\) (divided by 2! for the duplicate 2s).
2.  Subtract permutations starting with 0: fix 0 as first digit, permute the remaining 2,1,2: \(\frac{3!}{2!}=3\).
3.  Valid numbers: \(12-3=9\). We can also count directly: first digit 1 gives 3 numbers (1022,1202,1220); first digit 2 gives 6 numbers (2012,2021,2102,2120,2201,2210), total 9.
ANSWER 8: D
---
### Problem 9:
A number is divisible by 3 if the sum of its digits is divisible by 3.
1.  Total possible 3-digit numbers: permutations of 4 digits taken 3 at a time: \(P(4,3)=4*3*2=24\).
2.  List all 3-digit combinations and their sums:
    - {1,2,3}: sum 6 (divisible by 3), permutations: \(3!=6\)
    - {1,2,4}: sum 7 (not divisible)
    - {1,3,4}: sum 8 (not divisible)
    - {2,3,4}: sum 9 (divisible by 3), permutations: \(3!=6\)
3.  Favorable outcomes: \(6+6=12\). Probability: \(12/24=1/2\).
ANSWER 9: C
---
### Problem 10:
Let \(P\) = Pedro's time (lower time = better finish). Translate the statements:
1.  Nico is 11 minutes behind Pedro: \(N = P +11\)
2.  Olympia is 2 minutes ahead of Melina: \(O = M -2 \implies M = O +2\)
3.  Olympia is 3 minutes behind Pedro: \(O = P +3\)
4.  Olympia is 6 minutes ahead of Luke: \(O = L -6 \implies L = O +6\)
Substitute \(O=P+3\) into the others:
- \(M = (P+3)+2 = P+5\)
- \(L = (P+3)+6 = P+9\)
- \(N = P+11\)
Order from fastest (smallest time) to slowest:
1.  Pedro (\(P\))
2.  Olympia (\(P+3\))
3.  Melina (\(P+5\))
4.  Luke (\(P+9\))
5.  Nico (\(P+11\))
Fourth place is Luke.
ANSWER 10: A
---
### Problem 11:
Use cents to avoid decimals:
1.  Left side value: \(20*25 + 10*10 = 500 + 100 = 600\) cents.
2.  Right side value: \(10*25 + 10n = 250 +10n\) cents.
3.  Set equal: \(250 +10n =600 \implies 10n=350 \implies n=35\).
ANSWER 11: D
---
### Problem 12:
Let \(x\) = number of months until their goldfish counts are equal.
- Brent's goldfish after \(x\) months: \(4*(4^x) =4^{x+1}\) (quadruples monthly).
- Gretel's goldfish after \(x\) months: \(128*(2^x)\) (doubles