🔬 Model Lab

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D

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

---

**Problem 1:**

**What is asked:** Find the fewest number of times pressing the 1/x key returns 32 to itself.

**Solution:**
- Start: 32
- Press 1/x once: 1/32 ≈ 0.03125
- Press 1/x twice: 1/(1/32) = 32 ✓

The reciprocal of a reciprocal gives the original number back. So pressing 1/x twice returns to the original.

**Key insight:** We need the minimum number of presses. After 1 press, we have 1/32 (not 32). After 2 presses, we have 32 again.

ANSWER 1: B

---

**Problem 2:**

**What is asked:** Find the units digit of 19^19 + 99^99.

**Solution:**
For 19^19:
- Units digit of 19 is 9
- Pattern of units digits of powers of 9: 9¹=9, 9²=81 (units 1), 9³=729 (units 9), 9⁴ (units 1)...
- Powers of 9 alternate: odd powers end in 9, even powers end in 1
- 19 is odd, so 19^19 ends in 9

For 99^99:
- Units digit of 99 is 9
- By same logic, 99^99 ends in 9 (99 is odd)

Sum: 9 + 9 = 18, which has units digit 8

ANSWER 2: D

---

**Problem 3:**

**What is asked:** Find the percent decrease in price per box.

**Solution:**
- Last week: 4 boxes for $5, so price per box = 5/4 = $1.25
- This week: 5 boxes for $4, so price per box = 4/5 = $0.80

Percent decrease = (1.25 - 0.80)/1.25 × 100%
= 0.45/1.25 × 100%
= 36%

Closest to 35%.

ANSWER 3: B

---

**Problem 4:**

**What is asked:** In how many ways can Jo climb 6 stairs taking 1, 2, or 3 at a time?

**Solution:**
Let f(n) = number of ways to climb n stairs.
- f(1) = 1: {1}
- f(2) = 2: {1,1}, {2}
- f(3) = 4: {1,1,1}, {1,2}, {2,1}, {3}
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24

ANSWER 4: E

---

**Problem 5:**

**What is asked:** Which statement about o² + no is always true, where o is odd?

**Solution:**
Since o is odd, let o = 2k+1.
- o² = (2k+1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 (odd)
- no = n(2k+1) = 2nk + n

So o² + no = odd + (2nk + n) = odd + n

- If n is even: odd + even = odd
- If n is odd: odd + odd = even

This is **always even** is false. Let me reconsider.

Actually: o² + no = o(o + n)
Since o is odd and o² is odd:
- o² + no = o² + no
- Factor: o(o + n)

Since o is odd:
- If (o+n) is even, then o(o+n) is even
- If (o+n) is odd, then o(o+n) is odd

Since o is odd, (o+n) is even iff n is odd, and (o+n) is odd iff n is even.
So o(o+n) is even when n is odd, odd when n is even.

Wait, let me verify: o² + no = o(o+n). Since o is odd, the product is odd iff (o+n) is odd, which happens iff n is even.

ANSWER 5: E

---

**Problem 6:**

**What is asked:** Two integers are added to {3, 3, 8, 11, 28} to double the range while keeping mode and median unchanged. Find the maximum sum.

**Solution:**
- Original: 3, 3, 8, 11, 28
- Range = 28 - 3 = 25
- Mode = 3 (appears twice)
- Median of 5 numbers is the 3rd element = 8

New range must = 50. Current span is 25, so we need to extend it to 50.
- Could make minimum -22 (since 28 - (-22) = 50), or
- Could make maximum 53 (since 53 - 3 = 50)

To keep mode as 3: need 3 to appear most frequently. With two added numbers, 3 still appears twice, so the two new numbers cannot both equal any other single value.

To keep median = 8 with 7 numbers: the 4th element must be 8.
After adding two numbers to {3, 3, 8, 11, 28}, sorted list must have 8 in position 4.

To maximize sum, use 53 and another number. For median to be 8 with 7 elements, position 4 must be 8.
If we add 53 and x: sorted could be 3, 3, x, 8, 11, 28, 53 if 3 ≤ x ≤ 8.
Maximum x = 8, so sum = 53 + 8 = 61.

ANSWER 6: E

---

**Problem 7:**

**What is asked:** Find the smallest whole number s such that 6.5, 10, and s form a triangle.

**Solution:**
Triangle inequality: sum of any two sides > third side.
- 6.5 + s > 10 → s > 3.5
- 6.5 + 10 > s → s < 16.5
- s + 10 > 6.5 (always true for positive s)

So 3.5 < s < 16.5.
Smallest whole number is s = 4.

ANSWER 7: B

---

**Problem 8:**

**What is asked:** Original pasta length if Henry eats 3-inch pieces from the middle, ending with 10 pieces totaling 17 inches.

**Solution:**
Each bite from the middle of one piece splits it into two pieces, so each bite increases the piece count by 1.
- Start: 1 piece
- End: 10 pieces
- Number of bites: 9

Each bite removes 3 inches.
Total removed: 9 × 3 = 27 inches

Original length: 17 + 27 = 44 inches

ANSWER 8: D

---

**Problem 9:**

**What is asked:** Count 4-digit passwords (0-9) not starting with 911.

**Solution:**
- Total passwords: 10⁴ = 10,000
- Passwords starting with 911: 10 (the last digit can be 0-9)
- Valid passwords: 10,000 - 10 = 9,990

ANSWER 9: D

---

**Problem 10:**

**What is asked:** Find who sits in seat #2, given constraints and that each statement is false.

**Solution:**
Given: Bret is in seat #3.

False statements:
1. "Bret is next to Carl" → Carl is NOT adjacent to seat #3, so Carl is in seat #1
2. "Abby is between Bret and Carl" → Abby is NOT between seats #1 and #3

With Carl in #1 and Bret in #3, positions are: Carl(#1), ?(#2), Bret(#3), ?(#4)

Remaining: Abby and Dana. Abby cannot be between #1 and #3, so Abby cannot be in #2.
Therefore: Dana in #2, Abby in #4.

ANSWER 10: D

---

**Problem 11:**

**What is asked:** Count ways to write 60 as sum of consecutive odd positive integers.

**Solution:**
Consecutive odd integers starting at 2k+1 for n terms:
Sum = (2k+1) + (2k+3) + ... = n(2k+1+2k+2n-1)/2 = n(2k+n) = 60

So n(2k+n) = 60, where n ≥ 2, k ≥ 0.

Testing divisors of 60:
- n=2: 2k+2=30, k=14. Sum: 29+31=60 ✓
- n=3: 2k+3=20, k=8.5 (not integer)
- n=4: 2k+4=15, k=5.5 (not integer)
- n=5: 2k+5=12, k=3.5 (not integer)
- n=6: 2k+6=10, k=2. Sum: 5+7+9+11+13+15=60 ✓
- n=10: 2k+10=6, k=-2 (invalid)
- n=12: 2k+12=5, k=-3.5 (invalid)
- n=15: 2k+15=4, k=-5.5 (invalid)
- n=20: 2k+20=3, k=-8.5 (invalid)
- n=30: 2k+30=2, k=-14 (invalid)
- n=60: 2k+60=1, k=-29.5 (invalid)

Valid: n=2, n=6. That's 2 ways.

ANSWER 11: B

---

**Problem 12:**

**What is asked:** Fewest jumps for Greta to reach position 2023 (right +5, left -3)

Problem 1:

Let the displayed number be \(32\).

A reciprocal key replaces a number \(x\) by \(1/x\).

- After 1 press: \(32 \to \frac{1}{32}\)
- After 2 presses: \(\frac{1}{32} \to 32\)

So the display returns to 32 after 2 presses.

You cannot do it in 1 press, because the reciprocal of 32 is not 32.

ANSWER 1: B

Problem 2:

We only need the units digit of
\[
19^{19}+99^{99}.
\]

Look at the units digits of each power.

- \(19^{19}\) has the same units digit as \(9^{19}\).
  Powers of 9 alternate:
  \[
  9^1\to 9,\quad 9^2\to 1,\quad 9^3\to 9,\quad 9^4\to 1,\dots
  \]
  Since 19 is odd, \(9^{19}\) ends in 9.

- \(99^{99}\) has the same units digit as \(9^{99}\).
  Since 99 is odd, \(9^{99}\) also ends in 9.

Now add the units digits:
\[
9+9=18,
\]
so the units digit is 8.

ANSWER 2: D

Problem 3:

Original price: 4 boxes for \$5, so price per box is
\[
\frac{5}{4}=\$1.25.
\]

Sale price: 5 boxes for \$4, so price per box is
\[
\frac{4}{5}=\$0.80.
\]

Decrease in price per box:
\[
1.25-0.80=0.45.
\]

Percent decrease:
\[
\frac{0.45}{1.25}=0.36=36\%.
\]

The closest choice is \(35\%\).

ANSWER 3: B

Problem 4:

Let \(f(n)\) be the number of ways to climb \(n\) stairs using 1, 2, or 3 steps at a time.

To reach \(n\), the last step could be:

- 1 stair from \(n-1\)
- 2 stairs from \(n-2\)
- 3 stairs from \(n-3\)

So
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]

Compute:

- \(f(1)=1\)
- \(f(2)=2\)  \((1+1,\ 2)\)
- \(f(3)=4\)  \((1+1+1,\ 1+2,\ 2+1,\ 3)\)

Then:
\[
f(4)=4+2+1=7
\]
\[
f(5)=7+4+2=13
\]
\[
f(6)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E

Problem 5:

We are asked about
\[
o^2+no=o(o+n),
\]
where \(o\) is odd and \(n\) is any whole number.

Since \(o\) is odd:

- \(o^2\) is odd.
- \(no\) has the same parity as \(n\), because odd \(\times\) even = even and odd \(\times\) odd = odd.

So:

- if \(n\) is even, then \(no\) is even, and odd + even = odd
- if \(n\) is odd, then \(no\) is odd, and odd + odd = even

Thus the expression is odd only when \(n\) is even.

ANSWER 5: E

Problem 6:

The list is
\[
3,3,8,11,28.
\]

Current range:
\[
28-3=25.
\]

We insert two integers to double the range, so the new range must be
\[
2\cdot 25=50.
\]

So the new minimum and maximum must differ by 50.

The mode and median must remain unchanged.

- The mode is 3, since 3 appears twice. To keep 3 the unique mode, we should not add another 3.
- The median of the original 5-number list is 8. After adding 2 numbers, there will be 7 numbers, so the median is the 4th number. To keep the median 8, the 4th number must still be 8.

Because the original numbers are \(3,3,8,11,28\), to get range 50 with the largest possible sum, we want the maximum as large as possible and the minimum as small as possible, while keeping the median 8.

The smallest possible minimum is 3, and then the maximum must be
\[
3+50=53.
\]

Can we place the two new numbers so the median remains 8? Yes: choose numbers less than or equal to 8 so that the ordered list is
\[
3,3,8,8,11,28,53
\]
or similar, making the 4th number 8.

Then the added numbers are 8 and 53, with sum
\[
8+53=61.
\]

This is the maximum possible.

ANSWER 6: E

Problem 7:

The side lengths are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

For a triangle, the sum of any two sides must be greater than the third.

The two shorter sides are \(6.5\) and \(s\), so we need:
\[
6.5+s>10.
\]
Thus
\[
s>3.5.
\]

The smallest whole number greater than 3.5 is 4.

Check:
- \(6.5+4=10.5>10\)
- \(6.5+10>4\)
- \(10+4>6.5\)

So \(s=4\).

ANSWER 7: B

Problem 8:

Henry starts with one piece of pasta. Each bite:

- removes 3 inches from the middle of one piece,
- and therefore increases the number of pieces by 1.

He ends with 10 pieces, so he must have taken
\[
10-1=9
\]
bites.

Each bite removes 3 inches of pasta, so total pasta eaten is
\[
9\cdot 3=27.
\]

The final total length is 17 inches, so the original length was
\[
17+27=44.
\]

ANSWER 8: D

Problem 9:

A password has 4 digits, each from 0 to 9, so there are
\[
10^4=10000
\]
possible passwords total.

But passwords may not begin with 911.

How many begin with 911?

- First digit fixed: 9
- Second digit fixed: 1
- Third digit fixed: 1
- Fourth digit can be anything from 0 to 9: 10 choices

So there are
\[
10
\]
forbidden passwords.

Therefore the number allowed is
\[
10000-10=9990.
\]

ANSWER 9: D

Problem 10:

We have four people: Abby, Bret, Carl, Dana.

Joe says:

1. “Bret is next to Carl.”
2. “Abby is between Bret and Carl.”

Both statements are false.

Also, Bret is actually in seat #3.

We need who is in seat #2.

Since Bret is in seat 3, for statement 1 to be false, Carl is not in seat 2 or 4? More directly, Carl is not next to Bret. Since seat 3 is next to seats 2 and 4, Carl cannot be in 2 or 4. So Carl must be in seat 1.

Now seats 1 and 3 are Carl and Bret. The remaining seats 2 and 4 are Abby and Dana.

Statement 2 says Abby is between Bret and Carl. Since Bret and Carl are in seats 3 and 1, the seat between them would be seat 2. But the statement is false, so Abby is not in seat 2.

Therefore Abby must be in seat 4, and Dana in seat 2.

ANSWER 10: D

Problem 11:

We want the number of ways to write 60 as a sum of two or more consecutive odd positive integers.

Suppose we have \(k\) consecutive odd numbers starting at \(a\):
\[
a,\ a+2,\ a+4,\dots,a+2(k-1).
\]
Their sum is
\[
\frac{k}{2}\bigl(2a+2(k-1)\bigr)=k(a+k-1).
\]

So we need
\[
k(a+k-1)=60,
\]
where \(a\) is odd and positive.

We can check possible lengths \(k\) that divide 60.

Also note \(a=\frac{60}{k}-(k-1)\) must be a positive odd integer.

Try divisors \(k\ge 2\):

- \(k=2\): \(a=30-1=29\), odd positive → works:
  \[
  29+31=60
  \]
- \(k=3\): \(a=20-2=18\), even → no
- \(k=4\): \(a=15-3=12\), even → no
- \(k=5\): \(a=12-4=8\), even → no
- \(k=6\): \(a=10-5=5\), odd positive → works:
  \[
  5+7+9+11+13+15=60
  \]
- \(k=10\): \(a=6-9=-3\), not positive → no
- Larger \(k\) give nonpositive \(a\).

So there are exactly 2 ways.

ANSWER 11: B

Pro

Problem 1:  
We’re told a calculator’s reciprocal key replaces the current number by its reciprocal. If the display shows \(0.00032\) (as in “00032”), then after pressing \(1/x\) once, it becomes
\[
\frac{1}{0.00032} = \frac{1}{32\times 10^{-5}}=\frac{1}{32}\times 10^5.
\]
But we don’t need the exact value—just how many presses it takes to return to the original display.

Pressing \(1/x\) once does \(x \mapsto 1/x\).  
Pressing twice does
\[
x \mapsto \frac{1}{x} \mapsto \frac{1}{(1/x)}=x.
\]
So after 2 presses, the display returns to the starting number.

Since the question asks the *fewest positive* number of presses, the answer is 2.

Tempting wrong choice: 1 press would give \(1/x\), not \(x\), so it can’t be correct.

ANSWER 1: B


Problem 2:  
We want the units digit of \(19^{19}+99^{99}\).

Work mod 10.

- Units digit of \(19\) is \(9\). So \(19^{19}\) has units digit equal to \(9^{19}\) units digit.  
Powers of 9 cycle mod 10:  
\(9^1\to 9\), \(9^2\to 1\), \(9^3\to 9\), \(9^4\to 1\), … odd exponents give 9.  
Since \(19\) is odd, units digit of \(19^{19}\) is \(9\).

- Units digit of \(99\) is \(9\) as well, so units digit of \(99^{99}\) is \(9^{99}\).  
\(99\) is odd, so it also has units digit \(9\).

So the sum has units digit:
\[
9+9=18 \implies \text{units digit } 8.
\]

ANSWER 2: D


Problem 3:  
Original deal: 4 boxes for \$5.  
So original price per box:
\[
\frac{5}{4}=1.25\text{ dollars}.
\]

Sale deal: 5 boxes for \$4.  
New price per box:
\[
\frac{4}{5}=0.8\text{ dollars}.
\]

Percent decrease:
\[
\frac{1.25-0.8}{1.25}\times 100\%
= \frac{0.45}{1.25}\times 100\%.
\]
Compute:
\[
\frac{0.45}{1.25}=\frac{45/100}{125/100}=\frac{45}{125}=\frac{9}{25}=0.36.
\]
So percent decrease is \(0.36\times 100\%=36\%\).

Closest choice is 35%.

ANSWER 3: B


Problem 4:  
Jo climbs 6 stairs, taking 1, 2, or 3 at a time. We count the number of compositions of 6 using parts 1–3 in order.

Let \(f(n)\) = number of ways to climb \(n\) stairs.

\[
f(0)=1 \quad (\text{one way: do nothing})
\]
\[
f(1)=f(0)=1
\]
\[
f(2)=f(1)+f(0)=1+1=2
\]
\[
f(3)=f(2)+f(1)+f(0)=2+1+1=4
\]
\[
f(4)=f(3)+f(2)+f(1)=4+2+1=7
\]
\[
f(5)=f(4)+f(3)+f(2)=7+4+2=13
\]
\[
f(6)=f(5)+f(4)+f(3)=13+7+4=24
\]

So there are 24 ways.

ANSWER 4: E


Problem 5:  
Let \(o\) be odd. Let \(n\) be any whole number. Consider:
\[
o^2 + no = o(o+n).
\]

Since \(o\) is odd, the parity of the product depends on whether \(o+n\) is even or odd.

- If \(o+n\) is even, then \(o(o+n)\) is odd \(\times\) even = even.
- If \(o+n\) is odd, then it’s odd \(\times\) odd = odd.

Now, \(o+n\) is even exactly when \(n\) is odd?  
Because odd + odd = even, and odd + even = odd.  
So:
- \(n\) odd \(\Rightarrow o+n\) even \(\Rightarrow\) expression even.
- \(n\) even \(\Rightarrow o+n\) odd \(\Rightarrow\) expression odd.

Thus the expression is **odd only if \(n\) is even**.

That matches choice E.

ANSWER 5: E


Problem 6:  
We have the list: \(3, 3, 8, 11, 28\).  
The range doubles after inserting two integers, but mode and median stay unchanged.

Original sorted list has 5 numbers:
\[
\text{mode} = 3 \quad (\text{appears twice})
\]
\[
\text{median} = 8 \quad (\text{3rd number})
\]
\[
\text{range} = 28-3=25
\]
Doubling range means new range is:
\[
50.
\]

After adding 2 numbers, there will be 7 numbers total, so the median will be the 4th number. To keep median unchanged at 8, the 4th number must still be 8.

Currently, numbers below 8 are the two 3’s. If we add numbers, we must ensure we still have exactly 3 numbers \(\le 8\) before the 4th position equals 8. That means we **cannot** insert any new number less than 8, because that would increase how many numbers are below 8 and shift the median.

So the two inserted integers must be:
- \(\ge 8\)

Also, to keep the **mode** as 3, we cannot add any more 3’s (otherwise 3’s frequency changes) but adding numbers >3 won’t affect the mode unless we add a number that becomes the most frequent. With only two added numbers, the maximum frequency of any other value we create is at most 2; but mode currently is 2 occurrences of 3. If we add two equal numbers (say \(x,x\)), that would tie the mode at 2, not keep mode unchanged uniquely. The problem says the mode remains unchanged, which means 3 should still be the (unique) mode. So we should avoid making any number occur twice.

Thus best is to insert two distinct integers \( \ge 8\), neither equal to 3.

Now handle the range doubling:
New range = (new max) − (new min) = 50.

Since we cannot insert anything less than 3 (otherwise min decreases and median may shift; also median would be threatened), the smallest number likely remains 3. We also cannot insert another 3 (would alter mode frequency), so min stays 3.

So we need:
\[
\text{new max} = 3 + 50 = 53.
\]
Thus one inserted number should be 53.

The other inserted number should be as large as possible while still keeping the median 8. If we insert another large number, the max becomes larger than 53, which would increase the range beyond 50—unless we also lower the min, but we can’t.

So the max must be exactly 53. Therefore, the other inserted number must be \(\le 53\).

To maximize the sum of the two inserted numbers, we take:
\[
53 + 52 = 105
\]
But we must check median condition: inserting 52 (or anything between 8 and 53) does not create any new numbers below 8, so the 4th number remains 8. Distinct from 3, so mode remains 3.

Then the maximum possible sum is
\[
53+52=105.
\]
But answer choices are around 56–61, so clearly the question asks: “What is the maximum possible **sum of two additional numbers**?” Wait—choices are 56–61, not 105, meaning we must have misread: The original list range doubling implies new range is doubled, but if min might change by inserting smaller number, we can keep max less.

Let’s re-evaluate carefully: We are inserting two integers into the list to “double its range,” meaning:
\[
(\text{new max} - \text{new min}) = 2(28-3)=50.
\]
We are free to change the min by inserting numb

Problem 1:
The reciprocal key $1/x$ performs the operation $f(x) = 1/x$. If we press it once, we get $1/x$. If we press it twice, we get $1/(1/x) = x$. Thus, pressing the key twice returns the number to its original value. The question asks for the fewest *positive* number of times to return to the original value. Since 1 time results in $1/32$, we must press it 2 times.
ANSWER 1: B

Problem 2:
We need the units digit of $19^{19} + 99^{99}$.
The units digit of $19^{19}$ is the same as $9^{19}$. The powers of 9 follow a cycle: $9^1=9, 9^2=81, 9^3=729, \dots$ (units digits are 9, 1, 9, 1...). Since 19 is odd, the units digit of $9^{19}$ is 9.
The units digit of $99^{99}$ is the same as $9^{99}$. Since 99 is odd, the units digit is 9.
Summing them: $9 + 9 = 18$. The units digit is 8.
ANSWER 2: D

Problem 3:
Original price per box: $\$5 / 4 = \$1.25$.
New price per box: $\$4 / 5 = \$0.80$.
Decrease: $\$1.25 - \$0.80 = \$0.45$.
Percent decrease: $(0.45 / 1.25) \times 100\% = (45 / 125) \times 100\% = (9/25) \times 100\% = 36\%$.
The closest choice is 35%.
ANSWER 3: B

Problem 4:
Let $a_n$ be the number of ways to climb $n$ stairs.
$a_1 = 1$ (1)
$a_2 = 2$ (1+1, 2)
$a_3 = 4$ (1+1+1, 1+2, 2+1, 3)
For $n > 3$, $a_n = a_{n-1} + a_{n-2} + a_{n-3}$.
$a_4 = 4 + 2 + 1 = 7$
$a_5 = 7 + 4 + 2 = 13$
$a_6 = 13 + 7 + 4 = 24$
ANSWER 4: E

Problem 5:
$o$ is odd, so $o = 2k+1$. $o^2 = (2k+1)^2 = 4k^2 + 4k + 1$, which is odd.
$no$ is $n \times (\text{odd})$. If $n$ is even, $no$ is even. If $n$ is odd, $no$ is odd.
Expression: $o^2 + no$.
If $n$ is even: (odd) + (even) = odd.
If $n$ is odd: (odd) + (odd) = even.
The expression is even only if $n$ is odd. This doesn't match the choices perfectly, let's re-read. Wait, if $n$ is odd, $o^2 + no$ is even. If $n$ is even, $o^2 + no$ is odd.
Choice B says "always even", which is false. Choice A says "always odd", false.
Let's re-evaluate: $o(o+n)$. Since $o$ is odd, the parity depends on $(o+n)$.
If $n$ is odd, $o+n$ is even, so $o(o+n)$ is even.
If $n$ is even, $o+n$ is odd, so $o(o+n)$ is odd.
None of the choices seem to fit perfectly? Let's re-check the question. "o is an odd whole number". If $o=1, n=1$, $1+1=2$ (even). If $o=1, n=2$, $1+2=3$ (odd).
Actually, the question is simple: $o^2 + no = o(o+n)$. If $n$ is odd, $o+n$ is even, so the product is even. If $n$ is even, $o+n$ is odd, so the product is odd.
Wait, check choice B again. Maybe I misread. Let's look at the options. None seem correct. Let me re-read the prompt. Ah, perhaps $o^2+no$ is always even? No. Let's re-calculate. $1^2 + 1(1) = 2$ (even). $1^2 + 2(1) = 3$ (odd).
The problem might be flawed, but let's check if I missed a constraint. "o is an odd whole number". If $o=3, n=0$, $9+0=9$ (odd). If $o=3, n=1$, $9+3=12$ (even).
Actually, looking at the choices, maybe it's E? "odd only if n is even". Yes, that matches!
ANSWER 5: E

Problem 6:
List: 3, 3, 8, 11, 28. Range = $28-3 = 25$. Median = 8. Mode = 3.
New range must be 50. Let new numbers be $x, y$.
If we add $x, y$ such that the new max is 53 and min is 3, range is 50.
If we add 53 and 3, list: 3, 3, 3, 8, 11, 28, 53. Mode is 3, median is 8. Correct.
Sum = 56. Can we get higher? If we add 53 and 2, range is 51 (too big). If we add 53 and 4, range is 49 (too small).
ANSWER 6: A

Problem 7:
Triangle inequality: $10 - 6.5 < s < 10 + 6.5 \implies 3.5 < s < 16.5$.
Smallest whole number $s$ is 4.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches but creates a new piece (or splits one). Starting with 1 piece, each bite increases the number of pieces by 1. To have 10 pieces, he made 9 bites.
Total length removed = $9 \times 3 = 27$ inches.
Total length remaining = 17 inches.
Original length = $27 + 17 = 44$ inches.
ANSWER 8: D

Problem 9:
Total passwords = $10^4 = 10000$.
Passwords starting with 911: $9110, 9111, \dots, 9119$ (10 passwords).
Total valid = $10000 - 10 = 9990$.
ANSWER 9: D

Problem 10:
Bret is in #3. Statements: "Bret is next to Carl" (False $\implies$ Carl is not in #2 or #4). "Abby is between Bret and Carl" (False).
Since Bret is in #3, Carl cannot be in #2 or #4. Carl must be in #1.
Seats: 1: Carl, 2: ?, 3: Bret, 4: ?.
Remaining: Abby, Dana.
"Abby is between Bret and Carl" is false. If Abby were in #2, she would be between #1 (Carl) and #3 (Bret). This statement must be false, so Abby cannot be in #2.
Thus, Dana is in #2.
ANSWER 10: D

Problem 11:
Sum of $k$ consecutive odd integers starting at $a$: $a + (a+2) + \dots + (a+2(k-1)) = 60$.
$ka + 2(0+1+\dots+k-1) = ka + k(k-1) = 60$.
$k(a + k - 1) = 60$.
Since $k$ is the number of terms and terms are odd, $k$ must be a divisor of 60. Also, $a = (60/k) - k + 1$. Since $a$ must be odd, $(60/k) - k$ must be even.
If $k=2: a = 30 - 2 + 1 = 29$. (29, 31). Sum = 60.
If $k=3: a = 20 - 3 + 1 = 18$ (Even, reject).
If $k=4: a = 15 - 4 + 1 = 12$ (Even, reject).
If $k=5: a = 12 - 5 + 1 = 8$ (Even, reject).
If $k=6: a = 10 - 6 + 1 = 5$. (5, 7, 9, 11, 13, 15). Sum = 60.
Two ways.
ANSWER 11: B

Problem 12:
Let $x$ be jumps of +5, $y$ be jumps of -3.
$5x - 3y = 2023$. We want to minimize $x+y$.
$5x - 3y = 2023 \implies 5x \equiv 2023 \equiv 1 \pmod 3 \implies 2x \equiv 1 \equiv 4 \pmod 3 \implies x \equiv 2 \pmod 3$.
Let $x = 3k+2$.
$5(3k+2) - 3y = 2023 \implies 15k + 10 - 3y = 2023 \implies 3y = 15k - 2013 \implies y = 5k - 671$.
To minimize $x+y = (3k+2) + (5k-671) = 8k - 669$, we need the smallest $k$ such that $y \ge 0$.
$5k \ge 671 \implies k \ge 134.2$. So $k=135$.
$x = 3(135) + 2 = 405 + 2 = 407$.
$y = 5(135) - 671 = 675 - 671 = 4$.
Total jumps = $407 + 4 = 411$.
ANSWER 12: D

Problem 1:
The calculator's reciprocal function maps a displayed number \(x\) to \(1/x\). Starting from the displayed value 32, the first press yields \(1/32\). The second press then yields \(1/(1/32) = 32\). Thus the displayed value returns to 32 after exactly two presses. No smaller positive number of presses works, because a single press produces \(1/32 \neq 32\).

ANSWER 1: B

Problem 2:
Only the units digit of each power is needed. The units digits of successive powers of 19 cycle every 4 steps: 9, 1, 9, 1, …. Because the exponent 19 is congruent to 3 modulo 4, the units digit of \(19^{19}\) is 9. The units digits of successive powers of 99 are constantly 9, so the units digit of \(99^{99}\) is also 9. Adding these units digits gives a units digit of \(9+9=18\), which is 8.

ANSWER 2: D

Problem 3:
Originally four boxes cost $5, so one box costs \(5/4=1.25\) dollars. On sale five boxes cost $4, so one box costs \(4/5=0.8\) dollars. The absolute decrease per box is \(1.25-0.8=0.45\) dollars. The percent decrease relative to the original price is therefore \((0.45/1.25)\times100\%=36\%\), which rounds to the nearest choice 35 %.

ANSWER 3: B

Problem 4:
Let \(a_n\) be the number of ways to climb \(n\) stairs taking steps of 1, 2 or 3 at a time. The last step is either a single stair (preceded by a way to climb \(n-1\)), two stairs (preceded by a way to climb \(n-2\)), or three stairs (preceded by a way to climb \(n-3\)). Hence the recurrence \(a_n=a_{n-1}+a_{n-2}+a_{n-3}\) holds. The base cases are \(a_0=1\), \(a_1=1\), \(a_2=2\). Computing forward yields
\[
a_3=4,\quad a_4=7,\quad a_5=13,\quad a_6=24.
\]
Thus there are 24 ways for six stairs.

ANSWER 4: E

Problem 5:
Factor the expression: \(o^2+no=o(o+n)\). Because \(o\) is an odd whole number, the product is even precisely when the second factor \(o+n\) is even, i.e., when \(n\) is odd. Consequently the expression is odd precisely when \(n\) is even, which is statement E.

ANSWER 5: E

Problem 6:
The original list 3, 3, 8, 11, 28 already has median 8 and mode 3. Its range is \(28-3=25\). Inserting two integers \(x\le y\) must keep the median 8 and the mode 3, while doubling the range to 50. The new minimum must therefore be \(28-50=-22\). To preserve the mode 3, at least three 3’s are required, so one of the inserted numbers must be 3. The largest admissible second number that keeps the ordered list’s median equal to 8 is then 28, giving the pair \(-22,28\) whose sum is 6. Replacing the second number by 29 yields the admissible pair \(-22,29\) whose sum is 7, still preserving median 8 and mode 3. No larger sum is possible without either changing the median or destroying the uniqueness of the mode.

ANSWER 6: B

Problem 7:
The triangle inequality requires that the sum of any two sides exceeds the third. In particular
\[
6.5+10>s,\qquad 6.5+s>10,\qquad 10+s>6.5.
\]
The middle inequality simplifies to \(s>3.5\). The smallest whole number satisfying this strict inequality is therefore 4.

ANSWER 7: B

Problem 8:
Each bite removes one piece and replaces it by two pieces, thereby increasing the total number of pieces by 1. Starting with one piece and ending with ten pieces therefore requires exactly nine bites. Each bite also shortens the total length by 3 inches, so nine bites shorten the pasta by 27 inches. The final total length is 17 inches, hence the original length must have been \(17+27=44\) inches.

ANSWER 8: D

Problem 9:
There are \(10^4=10000\) four-digit strings (including leading zeros). Exactly 10 of them begin with the forbidden sequence 911. Subtracting yields \(10000-10=9990\) admissible passwords.

ANSWER 9: D

Problem 10:
Joe’s two statements are both false. Consequently Bret is not next to Carl, and Abby is not seated between Bret and Carl. Bret occupies seat 3, so the only seat that can possibly be next to Carl while satisfying the falsity conditions is seat 1. The only arrangement consistent with both statements being false is therefore Abby–Dana–Bret–Carl. Seat 2 is occupied by Dana.

ANSWER 10: D

Problem 11:
Suppose \(k\ge2\) consecutive odd positives sum to 60; their average is the middle term (or average of the two middle terms) and must equal \(60/k\). Hence \(k\) must be an odd divisor of 60 that is at least 3. The positive odd divisors of 60 are 1, 3, 5, 15. The admissible values \(k=3,5,15\) each produce a unique increasing sequence of odd positives:
\[
19+21+23=60,\qquad 7+9+11+13+15+5=60,\qquad 1+3+\dots+15=60.
\]
Thus exactly three representations exist.

ANSWER 11: C

Problem 12:
Each right jump of +5 and left jump of −3 changes position by an integer linear combination \(5a-3b\). The smallest nonnegative combination congruent to 2023 modulo \(\gcd(5,3)=1\) is required. Solving the linear Diophantine equation shows that the minimal total jumps occur when \(a=407\) and \(b=2\), giving a net displacement of \(5\cdot407-3\cdot2=2023\) after 409 jumps. No smaller nonnegative solution exists.

ANSWER 12: C

## Step 1: Problem 1 - Understand the given problem and the effect of the 1/x key.
The problem asks us to determine the fewest positive number of times the 1/x key must be depressed so that the display again reads 00032, starting from 00032.

## Step 2: Analyze the effect of pressing the 1/x key on the number 00032.
When 00032 is displayed and the 1/x key is pressed, the display becomes 1/32 = 0.03125.

## Step 3: Continue analyzing the effect of subsequent presses of the 1/x key.
Pressing the 1/x key again, we get 1/0.03125 = 32. So, after two presses, we are back to 32.

## Step 4: Determine the number of times the 1/x key needs to be pressed to return to 00032.
Since after 2 presses we are back to 32, the fewest positive number of times is 2.

## Step 5: Rule out the wrong choices for Problem 1.
Choices A, C, D, and E can be ruled out because we've found that 2 presses are sufficient.

## Step 6: Provide the answer for Problem 1.
ANSWER 1: B

## Step 7: Problem 2 - Understand what is being asked.
We need to find the units digit of $19^{19} + 99^{99}$.

## Step 8: Analyze the units digit of $19^{19}$.
The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. Since $9^1 = 9$, $9^2 = 81$, $9^3 = 729$, and $9^4 = 6561$, the pattern for the units digit of $9^n$ is 9, 1, 9, 1, ... for odd and even n respectively. Thus, $9^{19}$ has a units digit of 9.

## Step 9: Analyze the units digit of $99^{99}$.
Similarly, the units digit of $99^{99}$ is the same as the units digit of $9^{99}$, which is 9 because 99 is odd.

## Step 10: Calculate the units digit of $19^{19} + 99^{99}$.
The sum of the units digits is $9 + 9 = 18$, so the units digit is 8.

## Step 11: Rule out the wrong choices for Problem 2.
Choices A, B, C, and E can be ruled out because the units digit is 8.

## Step 12: Provide the answer for Problem 2.
ANSWER 2: D

## Step 13: Problem 3 - Calculate the original and sale price per box.
Originally, 4 boxes cost $5, so the price per box is $5/4 = $1.25. On sale, 5 boxes cost $4, so the price per box is $4/5 = $0.80.

## Step 14: Calculate the percent decrease in price per box.
The decrease is $1.25 - $0.80 = $0.45. The percent decrease is ($0.45 / $1.25) * 100 = 36%.

## Step 15: Rule out the wrong choices for Problem 3.
The closest percent decrease is 35% (since 36% is closer to 35% than to 40%).

## Step 16: Provide the answer for Problem 3.
ANSWER 3: B

## Step 17: Problem 4 - Understand the problem and the possible ways Jo can climb the stairs.
Jo can climb 1, 2, or 3 stairs at a time, and there are 6 stairs in total.

## Step 18: Break down the problem into smaller sub-problems to find the number of ways to climb 6 stairs.
Let's denote $a_n$ as the number of ways to climb n stairs. We have $a_1 = 1$, $a_2 = 2$, and $a_3 = 4$ because you can climb 1, 2, or 3 stairs in 1, 2, or 4 ways respectively.

## Step 19: Establish a recursive formula for $a_n$.
$a_n = a_{n-1} + a_{n-2} + a_{n-3}$ because you can reach the nth stair from the (n-1)th, (n-2)th, or (n-3)th stair.

## Step 20: Calculate $a_4$, $a_5$, and $a_6$ using the recursive formula.
$a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7$, $a_5 = a_4 + a_3 + a_2 = 7 + 4 + 2 = 13$, and $a_6 = a_5 + a_4 + a_3 = 13 + 7 + 4 = 24$.

## Step 21: Rule out the wrong choices for Problem 4.
Choices A, B, C, and D are incorrect because $a_6 = 24$.

## Step 22: Provide the answer for Problem 4.
ANSWER 4: E

## Step 23: Problem 5 - Analyze the given expression $o^2 + no$.
$o$ is odd, so $o^2$ is odd. $no$ is the product of $n$ (any whole number) and $o$ (odd), so $no$ is odd if $n$ is odd and even if $n$ is even.

## Step 24: Determine the parity of $o^2 + no$.
If $n$ is odd, $o^2 + no$ is odd + odd = even. If $n$ is even, $o^2 + no$ is odd + even = odd.

## Step 25: Evaluate the given choices based on the analysis.
The expression is even if $n$ is odd and odd if $n$ is even, so it's not always odd or always even. It is odd only if $n$ is even.

## Step 26: Provide the answer for Problem 5.
ANSWER 5: E

## Step 27: Problem 6 - Understand the initial list and the conditions.
The list is 3, 3, 8, 11, 28. The range is 28 - 3 = 25. The mode is 3, and the median is 8.

## Step 28: Determine the conditions for the new list after adding two integers.
The new range must be 50 (double the original range), the mode must remain 3, and the median must remain 8.

## Step 29: Analyze the possible values for the two new integers to maximize their sum while keeping the median and mode unchanged.
To keep the median 8, the new numbers must be such that one is greater than or equal to 8 and the other is less than or equal to 8, or both are greater than 11 (since the list will be 3, 3, 8, 11, 28, x, y and the median is the average of 8 and 11 when x and y are larger than 11).

## Step 30: Consider the range condition to maximize the sum.
The maximum value must be such that it doubles the range. So, if the minimum remains 3, the maximum must be 3 + 50 = 53.

## Step 31: Determine the two numbers that satisfy all conditions and maximize their sum.
To maximize the sum, one number should be as large as possible (53), and the other should be such that it doesn't change the median or mode. So, the larger number is 53, and to keep the median 8, the other number should be less than or equal to 8 but not 3 (to keep the mode 3). Thus, it could be 8 or less but not 3.

## Step 32: Provide the maximum possible sum.
If we choose 53 and a number less than 8 (but not 3 to avoid changing the mode), like 8 or 7 (since 3 is the mode and we can't have another number with the same frequency as 3), we get a sum. However, to maximize the sum while adhering to the conditions, we should consider numbers that fit the criteria. Choosing 53 and 8 (or any number less than 8 but not 3) gives a sum. The maximum sum is achieved with 53 and a number that is at most 8.

## Step 33: Calculate the maximum sum.
The maximum possible sum is $53 + 8 = 61$ (if we consider 8 as it's the median and 

Problem 1:
The reciprocal key computes 1/x. Starting with 32:
Press 1: 1/32 = 0.03125.
Press 2: 1/(1/32) = 32.
So the display returns to 32 after 2 presses.
ANSWER 1: B

Problem 2:
Units digit of 19^19: 19 ends in 9. Powers of 9 cycle: 9^1=9, 9^2=81 (1), odd exponent → 9. So units digit is 9.
Units digit of 99^99: 99 ends in 9, odd exponent → 9.
Sum: 9 + 9 = 18 → units digit 8.
ANSWER 2: D

Problem 3:
Original price per box: 5/4 = $1.25.
Sale price per box: 4/5 = $0.80.
Decrease: 1.25 − 0.80 = $0.45.
Percent decrease: (0.45 / 1.25) × 100% = 36%.
Closest choice is 35%.
ANSWER 3: B

Problem 4:
Number of ways to climb n stairs with steps 1, 2, or 3: f(0)=1, f(1)=1, f(2)=2, f(3)=4.
f(4)=4+2+1=7, f(5)=7+4+2=13, f(6)=13+7+4=24.
ANSWER 4: E

Problem 5:
o is odd, so o² is odd. no is even if n is even, odd if n is odd.
o² + no = odd + (even or odd).
If n even: odd + even = odd.
If n odd: odd + odd = even.
Thus the expression is odd only if n is even.
ANSWER 5: E

Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25. Double range = 50.
To keep min=3, max must be 53. Insert 53 and another integer y.
Median of 7 numbers must be 8 (4th number). So the three smallest must be ≤8 and the 4th is 8. Thus y ≤ 8.
Mode must remain 3. If y=8, 8 appears twice (tie with 3) → mode changes. If y=3, sum=56. If y=7, 3 appears twice, 7 once → mode 3, median 8. Sum = 53+7=60.
Larger sum not possible without breaking conditions.
ANSWER 6: D

Problem 7:
Triangle inequality: s + 6.5 > 10 ⇒ s > 3.5. Smallest whole number s is 4.
Check: 4+6.5=10.5 > 10, 4+10 > 6.5, 6.5+10 > 4. Valid.
ANSWER 7: B

Problem 8:
Each bite removes 3 inches and turns one piece into two, increasing piece count by 1.
Final pieces = 10 ⇒ 9 bites. Total eaten = 9×3 = 27 inches.
Final total length = 17 inches.
Original length = 17 + 27 = 44 inches.
ANSWER 8: D

Problem 9:
Total 4-digit passwords: 10⁴ = 10000.
Forbidden: starting with 9,1,1. Number of such passwords: 10 (any 4th digit).
Allowed: 10000 − 10 = 9990.
ANSWER 9: D

Problem 10:
Seats: #1, #2, #3(Bret), #4.
Joe's statements are false:
1. "Bret is next to Carl" false → Carl not in #2 or #4 → Carl in #1.
2. "Abby is between Bret and Carl" false → Abby not between them. With Carl in #1, Bret in #3, Abby would be between if in #2. So Abby not in #2 → Abby in #4.
Remaining seat #2 is Dana.
ANSWER 10: D

Problem 11:
Sum of k consecutive odd positive integers starting at a: k(a + k − 1) = 60, with a odd, k ≥ 2.
Let m = a + k − 1. Then k·m = 60, m ≥ k, and a = m − k + 1 odd ⇒ m and k have same parity.
Factor pairs (k,m) with k ≥ 2, m ≥ k:
(2,30): parity even/even → a=29 valid.
(3,20): odd/even → invalid.
(4,15): even/odd → invalid.
(5,12): odd/even → invalid.
(6,10): even/even → a=5 valid.
(10,6): m < k → a negative.
Total 2 ways.
ANSWER 11: B

Problem 12:
Let R = right jumps (+5), L = left jumps (−3). Net: 5R − 3L = 2023.
Minimize N = R + L.
5R ≡ 2023 (mod 3) → 2R ≡ 1 (mod 3) → R ≡ 2 (mod 3). Let R = 3k+2.
Then L = (5(3k+2)−2023)/3 = 5k − 671 ≥ 0 → k ≥ 135.
N = (3k+2) + (5k−671) = 8k − 669.
Minimum at k=135: N = 8×135 − 669 = 411.
ANSWER 12: D

# Problem 1

**What is asked:** Starting with 32 on the display, how few times must we press 1/x to see 32 again?

**Step-by-step:**
- Press 1: The display shows $1/32 = 0.03125$
- Press 2: The display shows $1/(1/32) = 32$

The reciprocal of the reciprocal returns the original number. Since we need a *positive* number of presses, the minimum is 2.

**Ruling out others:** 1 press gives $1/32$, not 32. So A is wrong. 2 is the smallest that works.

ANSWER 1: B

---

# Problem 2

**What is asked:** Find the units digit of $19^{19} + 99^{99}$.

**Step-by-step:**
- The units digit of a power depends only on the units digit of the base.
- Both 19 and 99 have units digit **9**.
- Powers of 9 cycle: $9^1 = 9$, $9^2 = 81$ (units 1), $9^3 = 729$ (units 9), $9^4$ (units 1), ...
- **Odd** exponents give units digit **9**; **even** exponents give units digit **1**.
- $19^{19}$: exponent 19 is odd → units digit = **9**
- $99^{99}$: exponent 99 is odd → units digit = **9**
- Sum of units digits: $9 + 9 = 18$ → units digit = **8**

ANSWER 2: D

---

# Problem 3

**What is asked:** Find the percent decrease in price per box.

**Step-by-step:**
- Last week: 4 boxes for \$5 → price per box = $\$5/4 = \$1.25$
- This week: 5 boxes for \$4 → price per box = $\$4/5 = \$0.80$
- Decrease = $\$1.25 - \$0.80 = \$0.45$
- Percent decrease = $\frac{0.45}{1.25} = 0.36 = 36\%$

36% is closest to **35%** among the choices.

ANSWER 3: B

---

# Problem 4

**What is asked:** How many ways can Jo climb 6 stairs, taking 1, 2, or 3 stairs at a time?

**Step-by-step:** Let $f(n)$ be the number of ways to climb $n$ stairs. The recurrence is $f(n) = f(n-1) + f(n-2) + f(n-3)$.

- $f(1) = 1$ (just "1")
- $f(2) = 2$ ("1+1" or "2")
- $f(3) = 4$ ("1+1+1", "1+2", "2+1", "3")
- $f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
- $f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
- $f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = 24$

ANSWER 4: E

---

# Problem 5

**What is asked:** Determine the parity of $o^2 + no$ where $o$ is odd and $n$ is any whole number.

**Step-by-step:**
Factor: $o^2 + no = o(o + n)$.

Since $o$ is odd:
- If $n$ is **even**: $o + n = \text{odd} + \text{even} = \text{odd}$. Then $o(o+n) = \text{odd} \times \text{odd} = \textbf{odd}$.
- If $n$ is **odd**: $o + n = \text{odd} + \text{odd} = \text{even}$. Then $o(o+n) = \text{odd} \times \text{even} = \textbf{even}$.

So the expression is odd **exactly when** $n$ is even.

Checking the choices:
- A: "always odd" — fails when $n$ is odd.
- B: "always even" — fails when $n$ is even.
- C: "even only if $n$ is even" — wrong; it's even when $n$ is *odd*.
- D: "odd only if $n$ is odd" — wrong; it's odd when $n$ is *even*.
- E: "odd only if $n$ is even" — correct! If the result is odd, then $n$ must be even.

ANSWER 5: E

---

# Problem 6

**What is asked:** Insert two integers into $\{3, 3, 8, 11, 28\}$ to double the range, keeping mode and median unchanged. Maximize their sum.

**Step-by-step:**
- Original: range = $28 - 3 = 25$, mode = 3, median = 8.
- New range must be $50$. New list has 7 numbers; median (4th value) must be 8.

To maximize the sum, we want the largest possible new maximum. Keep the minimum at 3, so new max = $3 + 50 = 53$.

For the median to stay 8, the 4th sorted value must be 8. We need one added number $\leq 8$ and one $> 8$ (which is 53).

To maximize the sum, make the smaller added number as large as possible: try $a = 8$. But then 8 appears twice (tied with 3), changing the mode. So $a = 7$.

New list: $3, 3, 7, 8, 11, 28, 53$
- Range: $53 - 3 = 50$ ✓
- Median (4th): $8$ ✓  
- Mode: $3$ (appears twice, all others once) ✓
- Sum: $7 + 53 = 60$

Trying to get 61 (e.g., $8 + 53$) breaks the mode. Other configurations give smaller sums.

ANSWER 6: D

---

# Problem 7

**What is asked:** Smallest whole number $s$ for a triangle with sides 6.5, 10, and $s$.

**Step-by-step:**
Triangle inequality requires:
1. $s + 6.5 > 10 \implies s > 3.5$
2. $s + 10 > 6.5 \implies s > -3.5$ (always true)
3. $6.5 + 10 > s \implies s < 16.5$

So $3.5 < s < 16.5$. The smallest whole number in this range is $s = 4$.

Check: sides 4, 6.5, 10 → $4 + 6.5 = 10.5 > 10$ ✓

ANSWER 7: B

---

# Problem 8

**What is asked:** Find the original length of the pasta.

**Step-by-step:**
- Start with 1 piece. Each bite splits one piece into two (eating 3 inches from the middle), adding 1 piece.
- End with 10 pieces → $10 - 1 = 9$ bites were taken.
- Total pasta eaten: $9 \times 3 = 27$ inches.
- Remaining pasta: 17 inches.
- Original length: $17 + 27 = 44$ inches.

ANSWER 8: D

---

# Problem 9

**What is asked:** How many 4-digit passwords don't begin with 9, 1, 1?

**Step-by-step:**
- Total 4-digit passwords (digits 0–9, repeats allowed): $10^4 = 10{,}000$.
- Passwords beginning with 9, 1, 1: the form is $9\text{-}1\text{-}1\text{-}\_$, where the last digit is any of 10 choices → $10$ passwords.
- Valid passwords: $10{,}000 - 10 = 9{,}990$.

ANSWER 9: D

---

# Problem 10

**What is asked:** Who is in seat #2?

**Step-by-step:**
Seats: #1, #2, #3, #4. Bret is in #3.

Joe's statements are both **false**:
1. "Bret is next to Carl" is FALSE → Carl is NOT in #2 or #4. So Carl is in **#1**.
2. "Abby is between Bret and Carl" is FALSE.

Remaining people: Abby and Dana for seats #2 and #4.

If Abby is in #2: she sits between Carl (#1) and Bret (#3), making statement 2 TRUE. But it must be FALSE. So Abby is NOT in #2.

Therefore: **Dana** is in #2, Abby is in #4.

Verification: #1 Carl, #2 Dana, #3 Bret, #4 Abby.
- Bret (#3) next to Carl (#1)? No → FALSE ✓
- Abby (#4) between Bret and Carl? No → FALSE ✓

ANSWER 10: D

---

# Problem 11

**What is asked:** How many ways can 60 be written as a sum of 2+ consecutive odd positive integers?

**Step-by-step:**
Let there be $k$ consecutive odd integers starting at odd $a$: $a, a+2, \ldots, a+2(k-1)$.

Sum $= k \cdot a + k(k-1) = k(a + k - 1) = 60$.

We need $k \geq 2$, $a \geq 1$ and odd. So $a + k - 1 \geq k$.

Factor pairs $(k,\; a+k-1)$ of 60 with $k \geq 2$:

 **Problem 1:**
The reciprocal key replaces a displayed number \(x\) with \(\frac1x\).  Applying it twice returns the original number because \(\frac1{1/x}=x\).  Starting from \(32\), one press gives \(\frac1{32}\neq 32\); two presses give \(32\) again.  Thus the smallest positive number of presses needed is \(2\).

ANSWER 1: B

---

**Problem 2:**
We need the units digit of \(19^{19}+99^{99}\).  Only the units digits of the bases matter, so this is the same as the units digit of \(9^{19}+9^{99}\).  The units digits of powers of \(9\) alternate \(9,1,9,1,\dots\); an odd exponent gives a units digit of \(9\).  Both exponents \(19\) and \(99\) are odd, so each term ends in \(9\).  Adding them: \(9+9=18\), so the final units digit is \(8\).

ANSWER 2: D

---

**Problem 3:**
Last week the price per box was \(\frac{5}{4}=1.25\) dollars.  This week it is \(\frac{4}{5}=0.80\) dollar.  The decrease is \(1.25-0.80=0.45\) dollar.  The percent decrease is
\[
\frac{0.45}{1.25}=0.36=36\%.
\]
\(36\%\) is \(1\%\) away from \(35\%\) and \(4\%\) away from \(40\%\), so it is closest to \(35\%\).

ANSWER 3: B

---

**Problem 4:**
Let \(f(n)\) be the number of ways to climb \(n\) stairs using steps of \(1\), \(2\), or \(3\).  Then \(f(0)=1\) and for \(n\ge 1\),
\[
f(n)=f(n-1)+f(n-2)+f(n-3).
\]
Compute:
\[
\begin{aligned}
f(1)&=1,\\
f(2)&=f(1)+f(0)=2,\\
f(3)&=f(2)+f(1)+f(0)=4,\\
f(4)&=f(3)+f(2)+f(1)=7,\\
f(5)&=f(4)+f(3)+f(2)=13,\\
f(6)&=f(5)+f(4)+f(3)=13+7+4=24.
\end{aligned}
\]
So there are \(24\) ways.

ANSWER 4: E

---

**Problem 5:**
Factor the expression:
\[
o^2+no=o(o+n).
\]
Since \(o\) is odd, the product is odd exactly when the second factor \((o+n)\) is odd.  Because \(o\) is odd, \(o+n\) is odd precisely when \(n\) is even.  Therefore the whole number is odd only if \(n\) is even.  (It is then automatically odd whenever \(n\) is even.)

ANSWER 5: E

---

**Problem 6:**
The original sorted list is \(3,3,8,11,28\).  Its range is \(28-3=25\), its mode is \(3\), and its median is \(8\).  After inserting two integers there are \(7\) numbers.

- **Range doubled:** new range must be \(50\).  Since \(3\) is present, the minimum is at least \(3\); to maximize the sum of the two new numbers we keep the minimum at \(3\), forcing the maximum to be \(3+50=53\).  Hence one added number must be \(53\).
- **Median unchanged:** the median of \(7\) numbers is the \(4\)th.  With two \(3\)s already, if both new numbers exceeded \(8\) the \(4\)th term would be greater than \(8\).  Thus at least one added number must be \(\le 8\).
- **Mode unchanged:** the mode must stay \(3\).  Adding another \(8\) would make \(8\) appear twice, tying the mode.  Adding a number equal to any existing non‑\(3\) value would also create a tie.  So the second added number should be a new integer less than \(8\).  The largest such integer is \(7\).

Check: the new list \(3,3,7,8,11,28,53\) has range \(50\), median \(8\), and unique mode \(3\).  The sum of the two added numbers is \(7+53=60\).

ANSWER 6: D

---

**Problem 7:**
For three lengths to form a triangle, the sum of any two sides must exceed the third.  The critical inequalities are
\[
s+6.5>10\quad\text{and}\quad 6.5+10>s.
\]
The first gives \(s>3.5\); the second gives \(s<16.5\).  Since \(s\) is a whole number, the smallest possible value is \(4\).

ANSWER 7: B

---

**Problem 8:**
Each bite removes \(3\) inches from a piece and splits that piece into two, increasing the total number of pieces by exactly \(1\).  Starting with \(1\) piece and ending with \(10\) pieces requires \(10-1=9\) bites.  Total pasta eaten is \(9\times 3=27\) inches.  Therefore the original length was
\[
17+27=44\text{ inches}.
\]

ANSWER 8: D

---

**Problem 9:**
There are \(10^4=10000\) possible four‑digit passwords with repetition allowed.  The only forbidden ones are those that begin \(9,1,1\); the fourth digit can be anything, giving \(10\) forbidden passwords.  Hence the number of allowable passwords is
\[
10000-10=9990.
\]

ANSWER 9: D

---

**Problem 10:**
Bret is in seat \(\#3\).

- Joe says “Bret is next to Carl,” which is false.  The seats next to \(\#3\) are \(\#2\) and \(\#4\), so Carl is **not** in either; thus Carl must be in seat \(\#1\).
- Joe says “Abby is between Bret and Carl,” which is false.  With Bret in \(\#3\) and Carl in \(\#1\), the only seat between them is \(\#2\).  Hence Abby is **not** in seat \(\#2\).

The remaining seats for Abby and Dana are \(\#2\) and \(\#4\).  Since Abby cannot be in \(\#2\), Dana must be sitting there.

ANSWER 10: D

---

**Problem 11:**
Let the first odd integer be \(a\) and let there be \(k\ge 2\) consecutive odd terms.  Their sum is
\[
a+(a+2)+\dots+(a+2(k-1))=k(a+k-1)=60.
\]
Thus \(k\) must divide \(60\), and \(a=\frac{60}{k}-k+1\) must be a positive odd integer.  Checking divisors \(k\ge 2\):

- \(k=2\): \(a=29\) (odd, good).
- \(k=3\): \(a=18\) (even, reject).
- \(k=4\): \(a=12\) (even, reject).
- \(k=5\): \(a=8\) (even, reject).
- \(k=6\): \(a=5\) (odd, good).
- \(k\ge 10\): \(a\le 0\) (reject).

Only \(k=2\) and \(k=6\) work, giving two representations.

ANSWER 11: B

---

**Problem 12:**
Let \(r\) be right jumps (\(+5\)) and \(l\) be left jumps (\(-3\)).  We need
\[
5r-3l=2023,
\]
and we want to minimize the total number of jumps \(r+l\).  Solving for \(r\),
\[
r=\frac{2023+3l}{5}.
\]
For \(r\) to be an integer, \(2023+3l\equiv 0\pmod 5\).  Since \(2023\equiv 3\pmod 5\), we need \(3l\equiv 2\pmod 5\), which gives \(l\equiv 4\pmod 5\).  The smallest non‑negative such \(l\) is \(4\), yielding
\[
r=\frac{2023+12}{5}=407.
\]
Thus \(r+l=407+4=411\).  Any other valid solution increases \(l\) by \(5\) and \(r\) by \(3\), raising the total by \(8\).  Hence the minimum number of jumps is \(411\).

ANSWER 12: D

# Problem 1:
We start with 32 on the display. Each press of 1/x replaces the current number with its reciprocal.

- Press 1: 32 → 1/32
- Press 2: 1/32 → 1/(1/32) = 32

After 2 presses, the display reads 00032 again. One press gives 1/32 ≠ 32, so 2 is the minimum.

ANSWER 1: B

---

# Problem 2:
We need the units digit of 19¹⁹ + 99⁹⁹.

**Units digit of 19¹⁹:** The units digit of 19 is 9. Powers of 9 cycle: 9, 1, 9, 1, ... (odd exponents → 9, even exponents → 1). Since 19 is odd, 19¹⁹ ends in **9**.

**Units digit of 99⁹⁹:** The units digit of 99 is 9. Same pattern applies. Since 99 is odd, 99⁹⁹ ends in **9**.

**Sum:** 9 + 9 = 18, so the units digit is **8**.

ANSWER 2: D

---

# Problem 3:
**Original price per box:** $5 ÷ 4 = $1.25

**Sale price per box:** $4 ÷ 5 = $0.80

**Percent decrease:** (1.25 − 0.80)/1.25 × 100 = 0.45/1.25 × 100 = 0.36 × 100 = 36%

This is closest to **35%**.

ANSWER 3: B

---

# Problem 4:
Let f(n) = number of ways to climb n stairs taking 1, 2, or 3 at a time.

- f(1) = 1
- f(2) = 2 (1+1, or 2)
- f(3) = 4 (1+1+1, 1+2, 2+1, 3)
- f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7
- f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13
- f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = **24**

ANSWER 4: E

---

# Problem 5:
We analyze o² + no = o(o + n) where o is odd.

- **o²** is odd (odd × odd = odd)
- **no** = n × o: this is odd if n is odd, even if n is even

**If n is even:** o² + no = odd + even = **odd** ✓
**If n is odd:** o² + no = odd + odd = **even** ✓

So the expression is odd only when n is even, and even only when n is odd.

This matches choice E: "it is odd only if n is even."

ANSWER 5: E

---

# Problem 6:
Original list: 3, 3, 8, 11, 28. Range = 25, Mode = 3, Median = 8.

New range must be 50. We add two integers; mode and median stay the same.

**Median stays 8:** With 7 numbers, the 4th (sorted) must be 8.

**Mode stays 3:** 3 must remain the most frequent value.

**Range = 50:** New max − new min = 50.

To **maximize** the sum of the two new numbers, we want the new maximum as large as possible. If we keep the minimum at 3, the new max = 3 + 50 = 53.

Adding 53 and checking: sorted list would be 3, 3, 8, 11, 28, 53 — but we need a 7th number. If we add another 3: sorted = 3, 3, 3, 8, 11, 28, 53. Median = 8 ✓, Mode = 3 (appears 3 times) ✓, Range = 50 ✓.

Sum = 3 + 53 = **56**.

(Adding any number less than 3 gives a smaller sum since the max would decrease by the same amount the min decreases.)

ANSWER 6: A

---

# Problem 7:
By the triangle inequality:
- 6.5 + 10 > s → s < 16.5
- 6.5 + s > 10 → s > 3.5
- 10 + s > 6.5 → always true for positive s

So 4 ≤ s ≤ 16. The smallest whole number value is **4**.

ANSWER 7: B

---

# Problem 8:
Each bite eats 3 inches from the middle of one piece, splitting it into two pieces (net +1 piece per bite).

Starting with 1 piece, after k bites there are k + 1 pieces.

Ending with 10 pieces: k + 1 = 10, so k = 9 bites.

Total pasta eaten: 9 × 3 = 27 inches.

Original length = remaining length + eaten length = 17 + 27 = **44 inches**.

ANSWER 8: D

---

# Problem 9:
Total 4-digit passwords (digits 0–9, repetitions allowed): 10⁴ = 10,000.

Passwords starting with 9, 1, 1: the first three digits are fixed as 9-1-1, and the fourth digit can be any of 0–9. That's 10 passwords.

Valid passwords = 10,000 − 10 = **9,990**.

ANSWER 9: D

---

# Problem 10:
Bret is in seat #3. Both of Joe's statements are false.

**Statement 1 is false:** "Bret is next to Carl" → Bret is NOT next to Carl. Seats next to #3 are #2 and #4, so Carl is NOT in #2 or #4. Carl must be in **#1**.

**Statement 2 is false:** "Abby is between Bret and Carl" → Abby is NOT between Bret (#3) and Carl (#1). The seat between #1 and #3 is #2, so Abby is NOT in #2.

Remaining people for seat #2: Not Bret (in #3), not Carl (in #1), not Abby. So **Dana** is in seat #2.

ANSWER 10: D

---

# Problem 11:
Consecutive odd integers starting from (2a+1) with k terms have sum = k(2a + k) = 60, where a ≥ 0, k ≥ 2.

Since 2a is even, 2a + k has the same parity as k. If k is odd, k(2a+k) is odd, but 60 is even. So **k must be even**.

- k = 2: 2(2a+2) = 60 → a = 14. Sequence: 29, 31. ✓
- k = 4: 4(2a+4) = 60 → 2a+4 = 15. Not integer. ✗
- k = 6: 6(2a+6) = 60 → a = 2. Sequence: 5, 7, 9, 11, 13, 15. ✓
- k = 8: 8(2a+8) = 60 → 2a+8 = 7.5. Not integer. ✗
- k = 10: 2a+10 = 6 → a = −2. Invalid. ✗

There are **2** ways.

ANSWER 11: B

---

# Problem 12:
Let r = right jumps, l = left jumps. We need 5r − 3l = 2023, minimizing r + l.

Solving: r = (2023 + 3l)/5. For r to be an integer, 2023 + 3l ≡ 0 (mod 5).

2023 ≡ 3 (mod 5), so 3 + 3l ≡ 0 (mod 5), giving 3l ≡ 2 (mod 5), so l ≡ 4 (mod 5).

Let l = 5k + 4 for k ≥ 0. Then r = (2023 + 15k + 12)/5 = 407 + 3k.

Total jumps = r + l = (407 + 3k) + (5k + 4) = 411 + 8k.

Minimum at k = 0: **411** jumps (with l = 4, r = 407).

Verification: 5(407) − 3(4) = 2035 − 12 = 2023 ✓

ANSWER 12: D

Problem 1:
The reciprocal key (1/x) changes a number to its reciprocal. If the display is 00032, pressing 1/x will change it to 1/32 = 0.03125, which is displayed as 003125 (assuming the display rounds or truncates to 6 digits, but the problem implies it shows the exact reciprocal, so we assume it shows 003125 or similar, but the exact display isn't critical). Pressing 1/x again returns to 32, which is 00032. Therefore, it takes 2 presses to return to 00032.

Tempting wrong choices: A (1 press would just give the reciprocal, not the original), C, D, E (more presses than necessary).

ANSWER 1: B

Problem 2:
To find the units digit of $19^{19} + 99^{99}$. The units digit of $19^{19}$ is the same as the units digit of $9^{19}$. The units digit of powers of 9 cycle every 2: 9, 1, 9, 1, ... Since 19 is odd, the units digit is 9. The units digit of $99^{99}$ is the same as $9^{99}$, and since 99 is odd, it's also 9. Adding them: 9 + 9 = 18, so the units digit is 8.

Tempting wrong choices: A (if you thought both were 0), B (if you thought both were 1), C (if you added the exponents or something), E (if you thought one was 9 and forgot to add).

ANSWER 2: E (units digit is 8? No, 9+9=18, units digit 8, but choice is E for 8? Wait, choices: A.0 B.1 C.2 D.8 E.9. So D is 8. I said 8, so D.
Correction: Units digit is 8, so answer D.

ANSWER 2: D

Problem 3:
Original price: 4 boxes for $5, so per box $5/4 = $1.25. New price: 5 boxes for $4, so per box $4/5 = $0.80. Decrease: $1.25 - $0.80 = $0.45. Percent decrease: ($0.45 / $1.25) * 100% = 36%. Closest to 35%.

Tempting wrong choices: A (30%, too low), C (40%, a bit high), D (45%, higher), E (65%, way off).

ANSWER 3: B

Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 steps at a time. Let's list the ways:
- All 1s: 1 way (1,1,1,1,1,1)
- Four 1s and one 2: positions for 2: 5 ways (since 6-2+1=5? No, number of ways to place one 2 in a sequence: for n stairs, but here stairs are climbed in steps, so it's the number of compositions.
Better to use recursion or list:
Let f(n) be ways to climb n stairs.
f(1) = 1 (1)
f(2) = 2 (1+1, 2)
f(3) = 4 (1+1+1, 1+2, 2+1, 3)
f(4) = f(3) + f(2) + f(1) = 4+2+1=7? No, for n=4: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 → 7 ways.
f(5) = f(4)+f(3)+f(2)=7+4+2=13
f(6)=f(5)+f(4)+f(3)=13+7+4=24

So 24 ways.

Tempting wrong choices: A (13 is f(5)), B (18? no), C (20? no), D (22? no).

ANSWER 4: E

Problem 5:
o is odd, n is any whole number. o² is odd (since odd*odd=odd). no is even if n even, odd if n odd. So o² + no: if n even, even + odd = odd? o² odd, no even (if n even), so odd + even = odd. If n odd, no odd, so odd + odd = even. So it's odd only if n is even? Let's see:
- n even: o² (odd) + even = odd
- n odd: odd + odd = even
So it's always odd when n even, and even when n odd. So "it is odd only if n is even" is correct, because if n even, it's odd; if n odd, it's even. So only when n even it's odd.

Choice E: "it is odd only if n is even" — yes.

Tempting: A (always odd? no, if n odd, even), B (always even? no), C (even only if n even? no, it's even when n odd), D (odd only if n odd? no, it's odd when n even).

ANSWER 5: E

Problem 6:
Original list: 3,3,8,11,28. Range: 28-3=25. Double range: 50. Mode: 3 (appears twice). Median: 8.
After adding two numbers, range should be 50, mode still 3, median still 8.
To maximize sum, we want large numbers. Range 50, so max - min =50. Min is still 3 (since mode is 3, and it must appear at least twice, so min can't be higher than 3? But we can add numbers larger than 28. If min is 3, max is 53. But if we set min to 3, max to 53, but original max is 28, so we add 53? But then the list is 3,3,8,11,28,53? But range is 50, yes. But mode: 3 still appears twice, but 53 once, so mode still 3? But we added only one 53, so yes. But we need to add two numbers. And median: after adding two numbers, total 7 numbers, median is the 4th when ordered. Original sorted: 3,3,8,11,28. Add two numbers, say x,y. To keep median 8, when sorted, the 4th element must be 8. So we need at least three numbers ≤8 and at least four numbers ≤ something? Total 7 numbers, sorted, 4th is median. Currently, without new numbers, the 4th is 11? Original: positions: 1:3,2:3,3:8,4:11,5:28. After adding two numbers, to have median 8, the 4th in sorted order must be 8. So we need at least four numbers ≤8? But currently only three numbers ≤8: 3,3,8. So we need to add at least one more number ≤8 to make four numbers ≤8. But if we add a number ≤8, but to maximize sum, we want large numbers, so perhaps add a number just below 8? But also, to keep min at 3, we cannot add below 3. But if we add a number above 28, min remains 3. But we need to add a number ≤8 to make the fourth position 8. Currently, if we add two large numbers, say 53 and 52, sorted: 3,3,8,11,28,52,53. Median is 4th: 11, not 8. To make median 8, we need the 4th element to be 8. So we need at least four numbers ≤8. Currently three: 3,3,8. So we need to add at least one more number ≤8. To maximize sum, we should add the largest possible number ≤8, which is 8. But if we add 8, then numbers ≤8: 3,3,8,8 — four of them. Then the sorted list: 3,3,8,8,11,28,x? We add two numbers: one is 8, and another large, say 53. Sorted: 3,3,8,8,11,28,53. Median is 4th: 8, good. Range: 53-3=50, good. Mode: 3 appears twice, 8 appears twice, so bimodal? But the problem says "the mode remains unchanged." Original mode was 3 (since it appeared twice, others once). Now both 3 and 8 appear twice, so mode is not unique. But the problem says "the mode remains unchanged," which might imply that 3 is still the only mode. So we cannot have another number appearing twice. So we cannot add another 8. We need to add a number ≤8 but not 3 or 8? But to maximize, we want large, but ≤8. If we add a number less than 8, but not 3, say 7. But then we add 7 and 53. Sorted: 3,3,7,8,11,28,53. Median is 4th: 8, good. Range 50. Mode: 3 appears twice, others once, so mode still 3. Good. Sum: 7+53=60. But is th

### Problem 1:
We start with the number 32, and the `1/x` key replaces the current number with its reciprocal. We need to find the minimum number of presses to return to 32.
- 1st press: `1/32` (reciprocal of 32)
- 2nd press: `1/(1/32) = 32` (reciprocal of `1/32` returns to the original number)
No fewer than 2 presses work. Wrong choices: A (1 press only gives the reciprocal, not the original), C/D/E (unnecessarily large).
ANSWER 1: B

---

### Problem 2:
We need the units digit of `19^19 + 99^99`. Only the units digit of the base and exponent affect the units digit of a power. For powers of 9, the units digit alternates: odd exponents give units digit 9, even exponents give 1.
- Both exponents 19 and 99 are odd, so each term has units digit 9.
- Sum of units digits: 9 + 9 = 18, so the total units digit is 8.
Wrong choices: A (would require sum of units digits ending in 0), B/C/E (do not match the calculation).
ANSWER 2: D

---

### Problem 3:
We calculate the percent decrease in price per box from last week ($5 for 4 boxes) to this week ($4 for 5 boxes).
- Last week price per box: `5/4 = $1.25`
- This week price per box: `4/5 = $0.80`
- Percent decrease = `(original - new)/original * 100 = (0.45/1.25)*100 = 36%`, which is closest to 35%.
Wrong choices: C (40% is further from 36% than 35%), A/D/E (far from the calculated 36%).
ANSWER 3: B

---

### Problem 4:
Jo climbs 6 stairs, taking 1, 2, or 3 stairs at a time. This is a tribonacci sequence problem, where `f(n)` = number of ways to climb n stairs, and `f(n) = f(n-1) + f(n-2) + f(n-3)` (the last step is 1, 2, or 3 stairs).
- Base cases: `f(1)=1, f(2)=2, f(3)=4`
- `f(4)=7, f(5)=13, f(6)=13+7+4=24`
Wrong choices: A/B/C/D (miscalculate the tribonacci sequence).
ANSWER 4: E

---

### Problem 5:
`o` is odd, `n` is any whole number. We analyze `o² + no = o(o + n)`. A product is odd only if both factors are odd. `o` is odd, so `o + n` must be odd for the product to be odd: odd + n is odd only if n is even. This means the number is odd only if n is even.
- A/B: Wrong, the number can be odd or even.
- C/D: Incorrectly state the conditions for even/odd results.
- E: Correctly matches the rule.
ANSWER 5: E

---

### Problem 6:
Original list: `[3,3,8,11,28]`. Original range = 25, so new range must be 50. Mode (3) and median (8) stay unchanged. To maximize the sum of the two added numbers, keep the minimum of the list at 3, so the new maximum is 3 + 50 = 53. The second added number can be at most 7 (to keep the median 8 and mode 3), sum = 7 + 53 = 60.
Wrong choices: E (61 would require adding 8, creating a bimodal list that changes the mode), A/B/C (lower sums).
ANSWER 6: D

---

### Problem 7:
Triangle sides 6.5, 10, s (s is a whole number). By the triangle inequality, the sum of the two smaller sides must exceed the largest side: `6.5 + s > 10 → s > 3.5`. The smallest whole number s is 4.
Wrong choices: A (3 is too small, 3+6.5=9.5 < 10), C/D/E (larger than the minimum).
ANSWER 7: B

---

### Problem 8:
Each bite splits 1 piece into 2, so each bite increases the number of pieces by 1. Starting with 1 piece, 10 pieces mean 9 bites. Each bite eats 3 inches, so total eaten = 9*3=27 inches. Remaining length is 17 inches, so original length = 27 +17=44 inches.
Wrong choices: A/B/C (miscalculate the number of bites), E (too large).
ANSWER 8: D

---

### Problem 9:
Total 4-digit passwords = 10^4=10000. Invalid passwords are those starting with `9,1,1`: there are 10 such passwords (the 4th digit can be any digit 0-9). Valid passwords = 10000 -10=9990.
Wrong choices: A/B/C (too small), E (does not subtract the 10 invalid passwords).
ANSWER 9: D

---

### Problem 10:
Bret is in seat 3. Joe's first statement is false: Bret is not next to Carl, so Carl cannot be in seat 2 or 4, so Carl is in seat 1. Joe's second statement is false: Abby cannot be between Bret (seat3) and Carl (seat1), so Abby cannot be in seat2. Seat2 must be Dana.
Wrong choices: A/B/C/E (do not fit the false statement conditions).
ANSWER 10: D

---

### Problem 11:
Sum of k consecutive odd integers starting at a: `k*(a +k-1)=60`, k≥2, a odd positive. Only two valid pairs exist:
1. k=2, a=29: 29+31=60
2. k=6, a=5: 5+7+9+11+13+15=60
Total of 2 ways.
Wrong choices: A/C/D/E (do not count only the valid sums).
ANSWER 11: B

---

### Problem 12:
Let x = number of +5 jumps, y=number of -3 jumps. `5x -3y=2023`, minimize x+y. Modulo 5: 3y ≡ -3 mod5 → y=5k-1. The smallest k=1 gives y=4, x=407. Total jumps=407+4=411.
Wrong choices: A/B/C/E (higher jump counts).
ANSWER 12: D

### Problem 1:
We need the fewest presses of the 1/x key to return to 32 (since 00032 = 32). Each press of 1/x takes the reciprocal of the current value. 
- 1st press: \( \frac{1}{32} \) (not 32)
- 2nd press: \( \frac{1}{1/32} = 32 \) (matches the original display)
Pressing twice returns to the original value, so the minimum number of presses is 2.
ANSWER 1: B
---
### Problem 2:
The units digit of a sum depends only on the units digits of the addends. 
- Units digit of \(19^{19}\) is the same as \(9^{19}\). Powers of 9 cycle every 2: \(9^1=9\) (units 9), \(9^2=81\) (units 1), so odd exponents of 9 have units digit 9. Thus \(9^{19}\) has units digit 9.
- Units digit of \(99^{99}\) is the same as \(9^{99}\). 99 is odd, so units digit is 9.
- Sum of units digits: \(9 + 9 = 18\), so the units digit of the total is 8.
ANSWER 2: D
---
### Problem 3:
Calculate the price per box before and during the sale:
- Original price: \( \frac{\$5}{4 \text{ boxes}} = \$1.25 \) per box.
- Sale price: \( \frac{\$4}{5 \text{ boxes}} = \$0.80 \) per box.
The price decrease is \(1.25 - 0.80 = \$0.45\). 
Percent decrease is \( \frac{\text{decrease}}{\text{original price}} \times 100 = \frac{0.45}{1.25} \times 100 = 36\% \). 36% is closest to 35%.
ANSWER 3: B
---
### Problem 4:
Let \(a_n\) = number of ways to climb \(n\) stairs with steps of 1, 2, or 3. 
- Base cases: \(a_0=1\) (1 way to climb 0 stairs), \(a_1=1\) (only 1 step), \(a_2=2\) (1+1 or 2), \(a_3=4\) (1+1+1, 1+2, 2+1, 3).
- For \(n \geq4\), the last step is 1, 2, or 3, so \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
Compute up to \(n=6\):
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 =7\)
- \(a_5 = a_4 + a_3 + a_2 =7 +4 +2=13\)
- \(a_6 = a_5 + a_4 + a_3 =13 +7 +4=24\)
ANSWER 4: E
---
### Problem 5:
Factor the expression: \(o^2 + no = o(o + n)\). Since \(o\) is odd:
- Odd × (any integer) has the same parity as the integer in parentheses.
- \(o\) is odd, so \(o + n\) is odd if \(n\) is even (odd + even = odd), and even if \(n\) is odd (odd + odd = even).
Thus \(o(o+n)\) is odd only when \(n\) is even.
ANSWER 5: E
---
### Problem 6:
Original list: [3,3,8,11,28]. Original range = \(28-3=25\), so new range = 50. Original mode = 3 (frequency 2), original median = 8 (3rd element of 5).
To maximize the sum of the two added numbers, we make the new maximum as large as possible: new max = \(3 + 50 = 53\) (keeping the original min 3, so range 50). 
To keep the median 8 (4th element of the new 7-element list), we need at least one added number ≤8 (so the 4th element remains 8). To keep the mode 3, we cannot add a number that creates a new mode (no other number can have frequency ≥2, unless it is 3). The largest possible number ≤8 that does not conflict with the mode is 7 (adding 7 does not increase the frequency of any number beyond 2, so mode remains 3).
The two added numbers are 7 and 53, sum = \(7 +53=60\). This satisfies all conditions: new list [3,3,7,8,11,28,53] has mode 3, median 8, range 50.
ANSWER 6: D
---
### Problem 7:
By the triangle inequality, the sum of any two sides must exceed the third:
1. \(6.5 + 10 > s \implies s <16.5\)
2. \(6.5 + s >10 \implies s>3.5\)
3. \(10 +s>6.5\) (always true for positive \(s\))
Since \(s\) is a whole number, the smallest possible \(s\) is 4.
ANSWER 7: B
---
### Problem 8:
Each bite removes 3 inches of pasta and increases the number of pieces by 1 (cutting a piece into two). Starting with 1 piece, ending with 10 pieces means there were \(10 -1=9\) bites. Total pasta eaten is \(9 \times3=27\) inches. The original length is the final total length plus eaten pasta: \(17 +27=44\).
ANSWER 8: D
---
### Problem 9:
Total unrestricted 4-digit passwords: \(10^4=10000\) (each digit 0-9). Forbidden passwords start with 9,1,1: the first three digits are fixed as 9,1,1, and the 4th digit can be any of 0-9, so 10 forbidden passwords. Total allowed: \(10000 -10=9990\).
ANSWER 9: D
---
### Problem 10:
Bret is in seat 3. Joe’s first statement ("Bret is next to Carl") is false, so Carl cannot be in seats 2 or 4 (adjacent to 3). Thus Carl is in seat 1.
Joe’s second statement ("Abby is between Bret and Carl") is false. Bret is in 3, Carl in 1: the only seat between them is seat 2. So Abby cannot be in seat 2. The remaining seat 2 must be Dana.
ANSWER 10: D
---
### Problem 11:
Let the sum of \(k\) consecutive odd positive integers (first term \(a\), odd) be 60. The sum is:
\[ S = k(a + k -1) =60 \]
Thus \(a = \frac{60}{k} -k +1\), where \(a\) is a positive odd integer, \(k\geq2\). \(k\) must divide 60. Test divisors of 60 ≥2:
- \(k=2\): \(a=30 -2 +1=29\) (odd, positive: valid, \(29+31=60\))
- \(k=3\): \(a=20-3+1=18\) (even: invalid)
- \(k=4\): \(a=15-4+1=12\) (even: invalid)
- \(k=5\): \(a=12-5+1=8\) (even: invalid)
- \(k=6\): \(a=10-6+1=5\) (odd, positive: valid, \(5+7+9+11+13+15=60\))
- \(k\geq10\): \(a\) is negative (invalid)
Only 2 valid ways.
ANSWER 11: B
---
### Problem 12:
Let \(x\) = number of +5 jumps, \(y\) = number of -3 jumps. Total displacement: \(5x -3y=2023\). Minimize total jumps \(N=x+y\).
From \(5x -3y=2023\), \(y=\frac{5x -2023}{3}\). For \(y\) to be a non-negative integer:
1. \(5x \geq2023 \implies x\geq405\) (since \(2023/5=404.6\))
2. \(5x \equiv2023 \pmod{3}\). \(5\equiv2\pmod3\), \(2023\equiv1\pmod3\), so \(2x\equiv1\pmod3 \implies x\equiv2\pmod3\).
The smallest \(x\geq405\) with \(x\equiv2\pmod3\) is \(x=407\) (405 mod3=0, 406 mod3=1, 407 mod3=2). Then \(y=\frac{5*407 -2023}{3}=\frac{2035-2023}{3}=4\). Total jumps \(N=407+4=411\).
ANSWER 12: D

Problem 1:
Reciprocal of 32 is 1/32; reciprocal again gives 32. So 2 presses returns to 32.

ANSWER 1: B

Problem 2:
Units digit of 19^19: 9^19. Powers of 9 cycle 9,1; odd exponent → 9.
Units digit of 99^99: 9^99, odd exponent → 9.
9+9=18, units digit 8.

ANSWER 2: D

Problem 3:
Last week: $5/4 = $1.25 per box. This week: $4/5 = $0.80 per box.
Decrease: 1.25−0.80 = 0.45. Percent: 0.45/1.25 = 0.36 = 36%, closest to 35%.

ANSWER 3: B

Problem 4:
Ways to climb n stairs with steps 1,2,3 = tribonacci.
f(1)=1, f(2)=2, f(3)=4, f(4)=7, f(5)=13, f(6)=24.

ANSWER 4: E

Problem 5:
o² + no = o(o + n). o is odd. If n even, o+n odd, product odd. If n odd, o+n even, product even.
So it's odd only when n is even.

ANSWER 5: E

Problem 6:
List: 3,3,8,11,28. Range = 28−3 = 25. Double range = 50.
To maximize sum, want largest possible new max. Keep minimum 3, so new max = 3+50 = 53.
Add 53 as one number. Need mode still 3 (appears most) and median unchanged.
Original median (5 numbers) = 8. With 7 numbers, median = 4th value.
Add 53 and another number x. To keep median 8 and mode 3.
Numbers: 3,3,8,11,28,53,x. Sort. For median (4th) = 8, need x ≥ 8 ideally. To maximize sum maximize x but x ≤ 53 keeping max 53 (can't exceed since range fixed at 50: max−min=50, min=3, max=53).
Set x as large as possible = 53? Then mode: 3 appears twice, 53 appears twice — tie, mode unchanged means mode stays 3 only. So can't have 53 twice. x can be anything ≤53 but not creating new mode.
To maximize, x close to 53 but distinct, say 53 not allowed twice. Use x giving max sum: pick x=53 forbidden. Next, we want sum 53+x max. Actually mode must remain 3 uniquely, so any value can appear at most once except 3 (twice). x just ≤53 and ≠53? Actually x=53 would tie mode. So second number ≤ ... maximize: choose another value not duplicating to give mode issue.

Let me reconsider: median must remain 8. Sorted seven: 3,3,8,...,28,53. The 4th element = 8 requires exactly: three elements ≤8 before position... Let's place: 3,3,8 are three smallest, then 11,28,53,x. For median (4th)=8, need 8 in position 4, so three values below or equal to 8 in first three: 3,3,8 — but 8 is the 3rd. Need 4th =8 means two 8's? 
Add x=8: list 3,3,8,8,11,28,53. Sorted, 4th=8. ✓ Mode: 3 twice, 8 twice → tie! Bad.
So median 8 with second number large is problematic.

Add both numbers ≥8 to keep 8 as median: then we'd have only one 8 and median would shift. Let's just compute: to keep median 8, the two added numbers should straddle so 8 stays 4th. Add one ≤8 and one ≥8: numbers a≤8, b≥8. Sorted seven, 4th element. We have 3,3,8,11,28 plus a,b. If a≤8: smallest set {3,3,a,8...}. 4th could be 8. Let a=3 (keeps mode 3 stronger): list 3,3,3,8,11,28,b. 4th=8 ✓. mode 3. b max = 53. Sum = 3+53=56.
Can we get higher? Add two large numbers? Then median moves up past 8. So 56.

ANSWER 6: A

Problem 7:
Triangle inequality: s + 6.5 > 10 → s > 3.5, and s > 10−6.5... also 6.5+10>s. Smallest whole: s=4.

ANSWER 7: B

Problem 8:
Each bite removes 3 inches and increases piece count by 1. Start 1 piece, end 10 pieces → 9 bites. Removed 9×3=27 inches. Remaining 17. Original = 17+27 = 44.

ANSWER 8: D

Problem 9:
Total: 10^4 = 10000. Forbidden: begin 911_, last digit free: 10 passwords. 10000−10 = 9990.

ANSWER 9: D

Problem 10:
Seats 1-4, Bret in seat 3.
Statement 1 false: Bret NOT next to Carl. Bret seat 3, neighbors 2,4. So Carl not in 2 or 4. Carl in seat 1.
Statement 2 false: Abby NOT between Bret and Carl. Bret seat3, Carl seat1, between is seat2. So Abby not in seat 2. Then Abby in seat 4, Dana in seat 2.

ANSWER 10: D

Problem 11:
Sum of consecutive odd integers starting from odd a, k terms: sum = k·(a + k−1) (since average = a+(k-1), as odds step 2: terms a, a+2,...; sum = k·a + 2·(0+1+...+(k-1)) = ka + k(k-1) = k(a+k-1)).
Set = 60.
k=2: 2(a+1)=60→a+1=30→a=29 odd ✓ (29,31). 
k=3: 3(a+2)=60→a+2=20→a=18 even ✗.
k=4: 4(a+3)=60→a+3=15→a=12 even ✗.
k=5: 5(a+4)=60→a+4=12→a=8 even ✗.
k=6: 6(a+5)=60→a+5=10→a=5 odd ✓ (5,7,9,11,13,15) sum=60 ✓.
k=8: 8(a+7)=60→ not integer.
k=10: 10(a+9)=60→a+9=6→a negative.
So 2 ways.

ANSWER 11: B

Problem 12:
Need net +2023 using +5 and −3. Let r right jumps, l left jumps: 5r−3l=2023, minimize r+l.
Minimize total. Express: we want small r+l. 5r−3l=2023.
r+l minimized. From 5r−3l=2023, l=(5r−2023)/3, need ≥0 and integer.
Total T=r+l=r+(5r−2023)/3=(8r−2023)/3. Minimize r (with constraints r+l, l≥0).
l≥0: 5r≥2023→r≥405 (405×5=2025). Need (5r−2023) divisible by 3.
r=405: 2025−2023=2, /3 no.
r=406: 2030−2023=7, no.
r=407: 2035−2023=12, l=4. T=407+4=411.
Check smaller T? T=(8r−2023)/3, increasing in r, so minimize r. r=407 first valid. T=411.

ANSWER 12: D

I'll solve each problem systematically.

---

**Problem 1:**
Taking the reciprocal of 32 gives 1/32. Taking the reciprocal again gives 32. So after 2 presses, we're back to 32.

ANSWER 1: B

---

**Problem 2:**
Units digit of 19^19: Powers of 9 cycle with period 2 (9,1,9,1...). 19 is odd, so units digit is **9**.
Units digit of 99^99: Same pattern. 99 is odd, so units digit is **9**.
9 + 9 = 18, units digit = **8**.

ANSWER 2: D

---

**Problem 3:**
Old price per box: $5/4 = $1.25
New price per box: $4/5 = $0.80
Decrease: $1.25 - $0.80 = $0.45
Percent decrease: 0.45/1.25 = 36%, closest to 35%.

ANSWER 3: B

---

**Problem 4:**
Let f(n) = ways to climb n stairs. f(1)=1, f(2)=2, f(3)=4, f(4)=f(3)+f(2)+f(1)=7, f(5)=f(4)+f(3)+f(2)=13, f(6)=f(5)+f(4)+f(3)=24.

ANSWER 4: E

---

**Problem 5:**
o²+no = o(o+n). Since o is odd, o(o+n) is odd when (o+n) is odd, i.e., when n is even; and even when n is odd. So it is odd only if n is even.

ANSWER 5: E

---

**Problem 6:**
Original range: 28-3=25. Double range = 50. New range must be 50.
Min stays at 3 (or lower), so max becomes 3+50=53, OR min decreases while max stays/increases.
Mode must remain 3 (so 3 must still appear most), median must remain 8 (middle of 7 values = 4th value).

To maximize the sum, make one number as large as possible: 3+50=53. The other number must keep median=8: sorted list with 53 and x must have 4th value=8. Adding 53 (large) and x: to keep median=8, we need x≤8. To keep mode=3, x≠anything that creates a new mode (x can be 8 since 8 appears once currently; if x=8, 8 appears twice, but 3 also appears twice — tie, mode changes). So x≤8 but x≠8. Maximum x=7 (not 3, to avoid issues; x=7 works). But wait, can x=3? That gives 3 appearing 3 times, mode still 3 ✓. Sum = 53+3=56? Let me try x=8: mode becomes tied — not valid. x=5: sum=58. Actually check x can be any value ≤8 excluding values that break mode. Max x=7 gives 53+7=60. Check: list is 3,3,3,7,8,11,28,53 — wait that's 8 numbers. Original 5 + 2 = 7 numbers. Sorted: 3,3,7,8,11,28,53. Median=8 ✓, mode=3 ✓, range=50 ✓. Sum=53+7=60.

Can x=8? Sorted: 3,3,8,8,11,28,53. Mode is 3 and 8 (bimodal) — mode changed. Invalid.
x=7 gives sum 60. Try making min smaller: say add -22 and 53: range=53-(-22)=75≠50. Need range exactly 50.

ANSWER 6: D

---

**Problem 7:**
Triangle inequality: sum of two sides > third side. 6.5+s>10 → s>3.5, so s≥4. Check: 4+6.5=10.5>10 ✓.

ANSWER 7: B

---

**Problem 8:**
Each bite creates one additional piece (splits one piece into two, removing 3 inches). Starting with 1 piece, after n bites: (1+n) pieces, length = L - 3n.
1+n=10 → n=9. Length = L-27=17 → L=44.

ANSWER 8: D

---

**Problem 9:**
Total passwords = 10^4 = 10000. Passwords beginning with 9,1,1: last digit has 10 choices → 10 forbidden passwords. Valid = 10000-10 = 9990.

ANSWER 9: D

---

**Problem 10:**
Both statements are false. "Bret is next to Carl" is FALSE, and "Abby is between Bret and Carl" is FALSE. Bret is in seat #3. Since Bret is NOT next to Carl, Carl ≠ #2 and Carl ≠ #4. So Carl = #1. Remaining seats #2 and #4 for Abby and Dana. "Abby is between Bret and Carl" is false. Carl=#1, Bret=#3: between them is #2. So Abby ≠ #2, meaning Abby=#4, Dana=#2.

ANSWER 10: D

---

**Problem 11:**
Consecutive odd integers: a, a+2, a+4,... for k terms. Sum = ka + k(k-1) = k(a+k-1) = 60. Need a odd, positive, k≥2. So k(a+k-1)=60, a=(60/k)-k+1 must be odd and positive.
- k=2: a=29 (odd✓) → 29+31=60 ✓
- k=3: a=19 (odd✓) → 19+21+23? Wait: a=60/3-2=18, not odd. ✗
- k=4: a=60/4-3=12, even ✗
- k=5: a=60/5-4=8, even ✗
- k=6: a=10-5=5 (odd✓) → 5+7+9+11+13+15=60 ✓
- k=10: a=6-9<0 ✗

Only 2 ways.

ANSWER 11: B

---

**Problem 12:**
Net right movement: 5R - 3L per combination. To get 2023: use 5a-3b=2023. Minimize a+b. 
2023 mod 8: 5a-3b=2023. Try b=1: 5a=2026, not divisible. b=3: 5a=2032, no. b=6: 5a=2041, no. b=9: 5a=2050=410×5, a=410. Total=419. Try a=406,b=1: 2030-3=2027≠. Notice 5×5-3×5=10, net 10 in 10 jumps. 2023=2020+3=404×5+3; 5×407-3×(405×... ). 5a-3b=2023, minimize a+b: a=(2023+3b)/5. Need 2023+3b≡0 mod5, 3+3b≡0, b≡4mod5. Min b=4: a=(2023+12)/5=407, total=411. Check b=9: a=408,total=417. So minimum is 407+4=411.

ANSWER 12: D