🔬 Model Lab

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take 

Problem 1:
We are asked for the average weight of all 10 children together.

Average = total weight divided by number of children.

Total weight of 6 boys: $6 \times 150 = 900$ pounds.
Total weight of 4 girls: $4 \times 120 = 480$ pounds.
Combined total: $900+480 = 1380$ pounds for 10 children.

Average = $1380 / 10 = 138$ pounds.

This is not simply $(150+120)/2 =135$, choice A, because there are more boys than girls, so the average is weighted toward the boys. It is between 120 and 150, closer to 150. 138 fits.
ANSWER 1: C

Problem 2:
We need the smallest 3-digit non-palindrome that is the sum of three distinct 2-digit palindromes.

A 2-digit palindrome has repeated digits: 11,22,33,44,55,66,77,88,99. Each is a multiple of 11. Hence the sum of three of them is also a multiple of 11.

So we need the least 3-digit multiple of 11, $\ge 100$, that can be written as such a sum, and check it is not a palindrome.

Multiples of 11: $\dots, 99, 110, 121, \dots$. There is no multiple of 11 from 100 to 109, since $9\times 11=99$ and $10\times 11=110$.

Consider 110. Can it be $11+22+77$, etc.? We need three distinct numbers from the list summing to 110. Dividing by 11, we need three distinct $k$ from 1 to 9 summing to $110/11=10$.
Indeed $1+2+7=10$, so $11+22+77=110$. They are distinct 2-digit palindromes.

Thus 110 is achievable, is 3-digit, and is not a palindrome read backwards: 011 is not 110.

Since no 100-109 number is a multiple of 11, no smaller 3-digit number can be such a sum. So $N=110$.

Sum of digits $=1+1+0=2$.
ANSWER 2: A

Problem 3:
Start with a $4$ by $4$ square.

Fold in half vertically: we get a double-thick $2$ wide by $4$ tall rectangle.

Cut both layers in half parallel to the fold, i.e. with a vertical cut in the middle of the 2-inch width. In terms of the original unfolded paper, this puts cuts at $x=1$ and $x=3$.

Unfolded, we get:
- middle strip $2$ wide by $4$ tall: the large rectangle.
- two side strips each $1$ wide by $4$ tall: the small rectangles.

Perimeter of one small: $2\times(1+4)=10$.
Perimeter of large: $2\times(2+4)=12$.

Ratio small to large $=10/12=5/6$.

Choices like 1/2 confuse area or width only; perimeter includes the long 4-inch sides which are common.
ANSWER 3: E

Problem 4:
We have TWO + TWO = FOUR, i.e. $2\times \text{TWO} = \text{FOUR}$.
$T=7$, $O$ even, all letters different digits.

Let TWO $=700+10W+O$. Twice it is between $1400$ and $1598$, so it is a 4-digit number starting with 1. Hence $F=1$.

Do column addition right to left:
Ones: $O+O = R$ plus carry $c1=0$ or $1$ to tens.
Tens: $W+W+c1 = U$ plus carry $c2=0$ or $1$ to hundreds.
Hundreds: $T+T+c2 = 7+7+c2 =14+c2$. This must equal $10F+O =10+O$.

Thus $14+c2 =10+O$, so $O =4+c2$. Since $c2$ is 0 or 1, $O=4$ or $5$. $O$ is even, so $O=4$ and $c2=0$.

Then $O+O=8$, so $R=8$, carry $c1=0$.

Tens: $W+W+0 = U$ with no carry, since $c2=0$. So $2W = U <10$. Thus $W\le 4$.

$W$ must differ from $T=7, F=1, O=4, R=8$.
So $W$ is 0,2, or 3 (not 1,4).

But $U=2W$ must also be distinct:
- If $W=0$, $U=0=W$, not allowed.
- If $W=2$, $U=4=O$, not allowed.
- If $W=3$, $U=6$, which is new and allowed.

So only $W=3$ works, giving $734+734=1468$.

Hence 0,1,2,4 are impossible.
ANSWER 4: D

Problem 5:
Tile $1$-$10$, die $1$-$6$, total $10\times 6=60$ equally likely pairs. Count pairs with product a perfect square.

Check each tile:
1: $1\times d$ square when $d=1,4$ → 2
2: $2,4,6,8,10,12$ → only 4 → 1
3: $3,6,9,12,15,18$ → only 9 → 1
4: $4,8,12,16,20,24$ → 4,16 → 2
5: $5,10,15,20,25,30$ → 25 → 1
6: $6,12,18,24,30,36$ → 36 → 1
7: $7,14,21,28,35,42$ → none → 0
8: $8,16,24,32,40,48$ → 16 → 1
9: $9,18,27,36,45,54$ → 9,36 → 2
10: $10,20,30,40,50,60$ → none → 0

Total favourable $=2+1+1+2+1+1+0+1+2+0=11$.

Probability $=11/60$.

This is slightly more than $1/6=10/60$, choice B, and slightly less than $1/5=12/60$, choice D.
ANSWER 5: C

Problem 6:
Let length $L$, width $W$, area $LW$.

New length $=L+20\%L =1.2L$.
New width $=W+50\%W =1.5W$.
New area $=1.2L \times 1.5W =1.8LW$.

That is $1.8$ times the old area, an increase of $0.8 =80\%$.

Simply adding $20\%+50\%=70\%$, choice C, is wrong because the increases multiply.
ANSWER 6: D

Problem 7:
The string repeats ABCDEDCBA ABCDEDCBA ...

Length of one block: A,B,C,D,E,D,C,B,A = 9 letters.

We want the 1992nd letter. Compute remainder mod 9.
$9\times 221 =1989$. $1992-1989=3$.

So it is the 3rd letter of the block, which is:
1:A, 2:B, 3:C. So C.

Choice E would be position 5, D position 4, etc.
ANSWER 7: C

Problem 8:
Each cube has 1 male snap and 5 female holes. To show only females, every male must be plugged into a female of another cube. So with $n$ cubes we need $n$ male-using connections.

A connected assembly of $n$ cubes needs at least $n-1$ face-adjacencies. Each adjacency can hide at most one male, because at an interface one side must be male and the other female; two males cannot meet.

So we need number of adjacencies $\ge n$.

$n=2$: Need 2 connections but two cubes can share only one face, impossible. One male would stay exposed.

$n=3$: Need 3 connections, so 3 adjacencies on 3 cubes, i.e. a triangle. Face-adjacency of cubes is on a cubic grid, which is bipartite and has no triangle. Also a connected 3 cubes can have at most 2 adjacencies in a path. So impossible.

$n=4$: Need 4 connections. Four cubes can form a 4-cycle, e.g. a $2\times2$ square. Each cube touches two neighbours, points its male to one neighbour and receives another's male in a female. This uses rotation to orient males around the loop. This is geometrically possible.

Thus smallest is 4, not 3.
ANSWER 8: B

Problem 9:
Let $f(n)$ be ways to climb $n$ stairs with steps 1,2,3.
$f(0)=1$, and for $n\ge1$, $f(n)=f(n-1)+f(n-2)+f(n-3)$, counting first step 1,2,3, with $f(\text{negative})=0$.

$f(1)=1$
$f(2)=f1+f0=1+1=2$
$f(3)=f2+f1+f0=2+1+1=4$
$f(4)=f3+f2+f1=4+2+1=7$
$f(5)=f4+f3+f2=7+4+2=13$
$f(6)=f5+f4+f3=13+7+4=24$.

So 24 ways.

This is tribonacci, not just Fibonacci for steps 1,2.
AN

Problem 1:
We are asked to find the average weight of all 10 children, given the average weights of the 6 boys and the 4 girls.

First, calculate the total weight of the boys:
$$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$

Next, calculate the total weight of the girls:
$$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$

Now, add the two totals to find the combined weight of all 10 children:
$$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$

Finally, divide by the total number of children (10):
$$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two averages: $\frac{150 + 120}{2} = 135$ pounds (Choice A), which fails to account for the fact that there are more boys than girls.

ANSWER 1: C

---

Problem 2:
We are looking for the sum of the digits of $N$, where $N$ is the smallest three-digit integer that is not a palindrome, but can be written as the sum of three distinct two-digit palindromes.

The two-digit palindromes are the two-digit multiples of $11$:
$$11, 22, 33, 44, 55, 66, 77, 88, 99$$

Since every two-digit palindrome is a multiple of $11$, the sum of any three of them must also be a multiple of $11$:
$$11a + 11b + 11c = 11(a + b + c)$$
where $1 \le a < b < c \le 9$.

Because $N$ must be a three-digit integer, the smallest possible three-digit multiple of $11$ is $110$.
- Is $110$ a palindrome? No, reading it backwards gives $011 = 11 \ne 110$.
- Can $110$ be written as the sum of three distinct two-digit palindromes? 
  $$\frac{110}{11} = 10$$
  We can write $10$ as the sum of three distinct single-digit integers, for example:
  $$1 + 2 + 7 = 10 \implies 11 + 22 + 77 = 110$$
  (Other valid combinations include $1+3+6$, $1+4+5$, and $2+3+5$).

Thus, the least such three-digit integer is $N = 110$.
The sum of the digits of $N$ is:
$$1 + 1 + 0 = 2$$

ANSWER 2: A

---

Problem 3:
A $4 \times 4$ inch square sheet of paper is folded in half vertically, creating a folded rectangle of width $2$ inches and height $4$ inches, with the crease along one vertical side.

Both layers are then cut in half parallel to the fold, which means a vertical cut is made $1$ inch from the crease:
1. The piece containing the crease unfolds into a single large rectangle of width $1 + 1 = 2$ inches and height $4$ inches.
2. The remaining piece consists of two separate small rectangular strips (one from each layer), each with width $1$ inch and height $4$ inches.

Now, calculate the perimeters:
- The perimeter of one small rectangle ($1 \times 4$ inches):
  $$\text{Perimeter}_{\text{small}} = 2 \times (1 + 4) = 10\text{ inches}$$
- The perimeter of the large rectangle ($2 \times 4$ inches):
  $$\text{Perimeter}_{\text{large}} = 2 \times (2 + 4) = 12\text{ inches}$$

The ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle is:
$$\frac{10}{12} = \frac{5}{6}$$

ANSWER 3: E

---

Problem 4:
We are given the cryptarithm:
$$\begin{array}{cccc}
& \text{T} & \text{W} & \text{O} \\
+ & \text{T} & \text{W} & \text{O} \\
\hline
\text{F} & \text{O} & \text{U} & \text{R}
\end{array}$$
with $T = 7$, each letter representing a distinct digit, and $O$ being an even number.

1. Substitute $T = 7$:
   $$\text{T} + \text{T} = 7 + 7 = 14$$
   This addition can produce at most a carry of $1$ from the tens column, so the hundreds sum is either $14$ or $15$. Therefore:
   $$\text{F} = 1 \quad \text{and} \quad \text{O} \in \{4, 5\}$$
2. Since $O$ is an even number, we must have:
   $$\text{O} = 4$$
3. Since the sum in the hundreds column is $14$ (which matches $\text{F}\text{O} = 14$), there is **no carry** from the tens column into the hundreds column.
4. Looking at the units column:
   $$\text{O} + \text{O} = 4 + 4 = 8 \implies \text{R} = 8$$
   with no carry into the tens column.
5. Looking at the tens column:
   $$\text{W} + \text{W} = \text{U}$$
   Because there is no carry to the hundreds column, $2W < 10$, which means $W \in \{0, 1, 2, 3, 4\}$.
   - If $W = 0$, then $U = 0$, but letters must be distinct ($W \ne U$).
   - If $W = 1$, this conflicts with $F = 1$.
   - If $W = 2$, then $U = 4$, which conflicts with $O = 4$.
   - If $W = 4$, this conflicts with $O = 4$.
   - If $W = 3$, then $U = 6$. 

Checking the assignment: $T=7, W=3, O=4, F=1, U=6, R=8$:
$$734 + 734 = 1468$$
All digits $\{1, 3, 4, 6, 7, 8\}$ are distinct. Thus, the only possible value for $W$ is $3$.

ANSWER 4: D

---

Problem 5:
There are $10$ tiles numbered $1$ to $10$ and a standard $6$-sided die numbered $1$ to $6$.
The total number of possible outcomes is:
$$10 \times 6 = 60$$

We want to find how many pairs $(T, D)$ produce a product $T \times D$ that is a perfect square:
- For $D = 1$: $T \times 1 = T$ is a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 2$: $T \times 2$ is a square $\implies T = 2k^2 \implies T \in \{2, 8\}$ ($2$ outcomes)
- For $D = 3$: $T \times 3$ is a square $\implies T = 3k^2 \implies T \in \{3\}$ ($1$ outcome)
- For $D = 4$: $T \times 4$ is a square $\implies T$ must be a square $\implies T \in \{1, 4, 9\}$ ($3$ outcomes)
- For $D = 5$: $T \times 5$ is a square $\implies T = 5k^2 \implies T \in \{5\}$ ($1$ outcome)
- For $D = 6$: $T \times 6$ is a square $\implies T = 6k^2 \implies T \in \{6\}$ ($1$ outcome)

Summing the favorable outcomes:
$$3 + 2 + 1 + 3 + 1 + 1 = 11\text{ outcomes}$$

The probability is:
$$\frac{11}{60}$$

ANSWER 5: C

---

Problem 6:
Let the original length be $L$ and the original width be $W$. The original area is:
$$A = L \times W$$

- Length is increased by $20\%$, so the new length is:
  $$L' = L \times (1 + 0.20) = 1.20L$$
- Width is increased by $50\%$, so the new width is:
  $$W' = W \times (1 + 0.50) = 1.50W$$

The new area is:
$$A' = L' \times W' = (1.20L) \times (1.50W) = 1.80 \times (L \times W) = 1.80A$$

The area has increased by:
$$1.80 - 1.00 = 0.80 = 80\%$$

Notice that adding the percentages ($20\% + 50\% = 70\%$, Choice C) is incorrect 

# Problem 1

**Find the combined average of 6 boys (avg 150 lbs) and 4 girls (avg 120 lbs).**

Total weight of boys: 6 × 150 = 900 pounds
Total weight of girls: 4 × 120 = 480 pounds
Total weight of all 10 children: 900 + 480 = 1380 pounds

Average = 1380 ÷ 10 = **138 pounds**

(Note: the answer is *not* (150+120)/2 = 135, because the groups have different sizes — the boys, being more numerous, pull the average up.)

ANSWER 1: C

# Problem 2

**Find the least three-digit non-palindrome that is a sum of three distinct two-digit palindromes.**

Two-digit palindromes are the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99.

We want the *smallest* sum that is at least 100. To keep the sum small, use the smallest values: 
- 11 + 22 + 66 = 99 (too small, only two digits)
- Next possible sums reaching 100: 11 + 22 + 77 = 110, or 11 + 33 + 66 = 110, or 22 + 33 + 55 = 110, or 11 + 44 + 55 = 110.

No combination of three distinct two-digit palindromes gives a three-digit number less than 110 (any sum of 100–109 would need a different combination, but checking shows 110 is the minimum achievable three-digit sum). 

Is 110 a palindrome? No — reading it backwards gives 011 = 11. ✓ So N = 110.

Sum of digits: 1 + 1 + 0 = **2**

ANSWER 2: A

# Problem 3

**Set up the geometry.** The 4×4 square is folded in half vertically, giving a 2-inch-wide, 4-inch-tall stack with the fold (crease) on one vertical edge.

Both layers are cut parallel to the fold — a vertical cut 1 inch from the crease, dividing the 2-inch-wide stack into two 1×4 pieces:
- The piece **containing the crease** unfolds into one large rectangle: **2 in × 4 in**.
- The piece **away from the crease** consists of two *separate* 1×4 layers, which unfold into **two small rectangles, each 1 in × 4 in**.

That's exactly "a large one and two small ones." ✓

**Compute perimeters:**
- Small rectangle (1 × 4): P = 2(1 + 4) = 10 inches
- Large rectangle (2 × 4): P = 2(2 + 4) = 12 inches

Ratio = 10/12 = **5/6**

ANSWER 3: E

# Problem 4

**Solve TWO + TWO = FOUR with T = 7 and O even.**

Write the columns (units → hundreds):
- Units: O + O = R + 10·c₁
- Tens: W + W + c₁ = U + 10·c₂
- Hundreds: T + T + c₂ = 10F + O

**Hundreds column:** 7 + 7 + c₂ = 14 + c₂ = 10F + O. Since F = 1, we need 10 + O = 14 + c₂, so O = 4 + c₂.
- If c₂ = 1: O = 5 (odd — rejected)
- If c₂ = 0: O = 4 ✓ (even), F = 1

**Units column:** 4 + 4 = 8, so R = 8, c₁ = 0.

**Tens column:** 2W + 0 = U with no carry (c₂ = 0), so 2W ≤ 9, meaning W ≤ 4. Test each candidate with all digits distinct (used so far: T=7, O=4, F=1, R=8):
- W = 0: U = 0 = W ✗
- W = 1: = F ✗
- W = 2: U = 4 = O ✗
- W = 3: U = 6 — all digits 7,3,4,1,6,8 distinct ✓
- W = 4: = O ✗

Check: 734 + 734 = 1468. ✓ The only possible value is W = **3**.

ANSWER 4: D

# Problem 5

**Count pairs (tile t ∈ 1–10, die d ∈ 1–6) with t·d a perfect square.** There are 60 equally likely pairs.

Go tile by tile:
- t = 1: d = 1, 4 → 2 pairs (1, 4)
- t = 2: 2d square → d = 2 (gives 4) → 1 pair
- t = 3: d = 3 (gives 9) → 1 pair
- t = 4: d = 1, 4 (gives 4, 16) → 2 pairs
- t = 5: d = 5 (gives 25) → 1 pair
- t = 6: d = 6 (gives 36) → 1 pair
- t = 7: 7d square needs d = 7·k² > 6 → 0 pairs
- t = 8: d = 2 (gives 16) → 1 pair
- t = 9: d = 1, 4 (gives 9, 36) → 2 pairs
- t = 10: 10d square needs d = 10·k² > 6 → 0 pairs

Total favorable: 2+1+1+2+1+1+0+1+2+0 = **11**

Probability = 11/60

ANSWER 5: C

# Problem 6

**Let the rectangle be L × W with area LW.**

New length: 1.2L. New width: 1.5W.
New area: (1.2L)(1.5W) = 1.8·LW.

The area is multiplied by 1.8, i.e., increased by **80%**.

(A tempting mistake is adding the percentages: 20% + 50% = 70% — that ignores the cross term 0.2 × 0.5 = 0.1 = 10% that makes it 80%.)

ANSWER 6: D

# Problem 7

**Find the 1992nd letter of ABCDEDCBAABCDEDCBA…**

The repeating block is "ABCDEDCBA", which has **9 letters**.

1992 ÷ 9: 9 × 221 = 1989, remainder 1992 − 1989 = **3**.

So the 1992nd letter is the 3rd letter of the block "ABCDEDCBA", which is **C**.

ANSWER 7: C

# Problem 8

**Each cube has 1 snap face and 5 receptacle faces. A face-to-face joint hides 2 faces and uses exactly 1 snap (inserted into 1 receptacle).**

Every cube's snap must be hidden inside a joint, and each joint consumes only one snap. With n cubes we have n snaps, so we need **n joints**. A connected arrangement of n cubes with n joints is a graph with n vertices and n edges — it must contain a cycle.

**Can 3 cubes work?** Three cubes would need 3 joints: A–B, B–C, and C–A, i.e., all three cubes mutually face-adjacent. Two cubes attached to opposite sides of a third can never touch each other face-to-face, so a 3-cycle of cubes is geometrically impossible. ✗

**Can 4 cubes work?** Arrange 4 cubes in a 2×2 square block. The adjacency graph is a 4-cycle: going around the block, each cube's snap inserts into the next cube's receptacle. Each cube's single snap is thus hidden, and every exposed face is a receptacle. ✓

The smallest number is **4**.

ANSWER 8: B

# Problem 9

**Count the ways to climb 6 stairs using steps of 1, 2, or 3.**

Let W(n) = number of ways to climb n stairs. The first move is 1, 2, or 3 stairs, so:
W(n) = W(n−1) + W(n−2) + W(n−3), with W(0) = 1 (do nothing), W(1) = 1, W(2) = 2 (1+1, 2).

Compute:
- W(3) = 2 + 1 + 1 = 4
- W(4) = 4 + 2 + 1 = 7
- W(5) = 7 + 4 + 2 = 13
- W(6) = 13 + 7 + 4 = **24**

ANSWER 9: E

# Problem 10

**Annie runs 25% faster than Bonnie, so their speeds are in the ratio 5 : 4.**

Annie first passes Bonnie when she has gained exactly one full lap (400 m) on her.

Since they run for the same time t, their distances are in the ratio 5 : 4. Annie's gain is the difference: 
Annie's distance − Bonnie's distance = 400 m.

If Annie runs 5k meters, Bonnie runs 4k meters, and 5k − 4k = k = 400 m. 
So Annie runs 5 × 400 = 2000 meters.

Laps for Annie: 2000 ÷ 400 = **5 laps**.

(At that moment Bonnie has run 4 laps — Annie has gained exactly one lap, as required.)

ANSWER 10: D

# Problem 1:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = **138 pounds**.

(The tempting wrong answer is 135, the average of 150 and 120 — but that would only be right if there were equal numbers of boys and girls. Since there are more boys, the average must be closer to 150 than to 120... wait, closer means above 135, which 138 is.)

ANSWER 1: C

# Problem 2:
Two-digit palindromes are 11, 22, 33, …, 99 — all multiples of 11.
So any sum of three of them is a multiple of 11: sum = 11(a + b + c) with a, b, c distinct digits from 1–9.

We want the least three-digit such sum that is **not** a palindrome.
- The smallest three-digit multiple of 11 is 110, needing a + b + c = 10 with distinct digits: e.g. 1 + 2 + 7, so 11 + 22 + 77 = 110. ✓
- Is 110 a palindrome? Reversed it is 011, so no. ✓

So N = 110, and the digit sum is 1 + 1 + 0 = **2**.

(121 is achievable too, but it's a palindrome and also larger.)

ANSWER 2: A

# Problem 3:
Start with a 4 × 4 square; fold in half vertically → a 2 × 4 double-layered rectangle (fold along one vertical edge).

Now cut both layers in half **parallel to the fold** (a vertical cut), giving two 1 × 4 double pieces:
- The piece containing the fold unfolds into a 2 × 4 rectangle (the large one).
- The other double piece separates into two 1 × 4 rectangles (the small ones).

Perimeters: small = 2(1 + 4) = 10; large = 2(2 + 4) = 12.

Ratio = 10/12 = **5/6**.

ANSWER 3: E

# Problem 4:
We need 2 × (TWO) = FOUR, with T = 7 and O even.

TWO = 700 + 10W + O, so
2(700 + 10W + O) = 1400 + 20W + 2O.

Since TWO is between 700 and 799, FOUR is between 1400 and 1598, so F = 1.
Then: 1400 + 20W + 2O = 1000 + 100·O + 10U + R, i.e.
400 + 20W + 2O = 100·O + 10U + R.

Test even O:
- O = 0: R = 0 too, but O and R must differ. ✗
- O = 2: 404 + 20W = 200 + 10U + R → 204 + 20W = 10U + R, impossible (right side ≤ 99). ✗
- O = 6: 20W = 384 + …, needs W ≥ 10. ✗ O = 8: even worse. ✗
- O = 4: 408 + 20W = 400 + 10U + R → 20W + 8 = 10U + R.
 - W = 0 → U = 0 (repeat), W = 1 → F = 1 (repeat), W = 2 → U = 4 = O (repeat),
 - **W = 3** → 68 → U = 6, R = 8: digits 7, 3, 4, 1, 6, 8 all distinct. ✓

Check: 734 + 734 = 1468 ✓ (F=1, O=4, U=6, R=8).

So W = **3**.

ANSWER 4: D

# Problem 5:
There are 10 × 6 = 60 equally likely (tile, die) outcomes. Count those whose product is a perfect square:

- Tile 1: die 1 (1), die 4 (4) → 2
- Tile 2: die 2 (4) → 1
- Tile 3: die 3 (9) → 1
- Tile 4: die 1 (4), die 4 (16) → 2
- Tile 5: die 5 (25) → 1
- Tile 6: die 6 (36) → 1
- Tile 7: none → 0
- Tile 8: die 2 (16) → 1
- Tile 9: die 1 (9), die 4 (36) → 2
- Tile 10: none → 0

Total = 11 favorable outcomes, so probability = **11/60**.

ANSWER 5: C

# Problem 6:
New length = 1.20L, new width = 1.50W.
New area = 1.20 × 1.50 × LW = 1.80·LW.

That is 180% of the original, an increase of **80%**.

(30% is the tempting error from adding 20% + 50% ; that's not how area scales.)

ANSWER 6: D

# Problem 7:
The block "ABCDEDCBA" has 9 letters and repeats forever.

1992 ÷ 9 = 221 remainder 3 (since 9 × 221 = 1989).

So the 1992nd letter is the 3rd letter of the block: A, B, **C**.

ANSWER 7: C

# Problem 8:
Each cube has 1 snap and 5 holes. To hide all snaps, every snap must be plugged into another cube's hole.

- With 2 or 3 cubes in a row/chain, some snaps end up on the outside (a "chain" always leaves the end cube's snap exposed unless it loops, and 3 cubes can't loop with this geometry).
- With 4 cubes: place two pairs. Snap two cubes together face-to-face (their snaps... ) — the standard construction is a 2 × 2 arrangement where each cube's snap points into a neighbor's hole, forming a cycle of 4. Every snap is covered, and all exposed faces are holes.

So the smallest number is **4**.

ANSWER 8: B

# Problem 9:
Let f(n) = number of ways to climb n stairs using steps of 1, 2, or 3. Then f(n) = f(n−1) + f(n−2) + f(n−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 9: E

# Problem 10:
Annie "first passes" Bonnie when she has gained exactly one full lap (400 m) on her.

Let Bonnie's speed be v; Annie's is 1.25v. Gain rate = 0.25v.
Time to gain 400 m: t = 400 / (0.25v) = 1600/v.

In that time Annie runs 1.25v × (1600/v) = 2000 meters = 2000/400 = **5 laps** (Bonnie has run 4).

(Choice A, 1¼ laps, only counts distance ratio, not the full lap of separation needed to lap her.)

ANSWER 10: D

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\), so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\), so their total weight is

\[
4\cdot 120=480
\]

Thus the total weight of all 10 children is

\[
900+480=1380
\]

Their average weight is

\[
\frac{1380}{10}=138
\]

A tempting incorrect choice is \(135\), which comes from averaging \(150\) and \(120\). That would only work if there were equal numbers of boys and girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

We need the least three-digit number that is:

1. not a palindrome, and  
2. the sum of three distinct two-digit palindromes.

The two-digit palindromes are

\[
11,22,33,44,55,66,77,88,99.
\]

Each is a multiple of \(11\). Therefore, the sum of any three of them is also a multiple of \(11\).

The smallest possible sum of three distinct ones is

\[
11+22+33=66,
\]

which is not three-digit.

To make a three-digit number, we need at least \(100\). The smallest multiple of \(11\) that is at least \(100\) is

\[
110.
\]

Can we make \(110\) from three distinct two-digit palindromes? Yes:

\[
11+22+77=110.
\]

The number \(110\) is not a palindrome, since reading it backward gives \(011\), not \(110\).

The sum of its digits is

\[
1+1+0=2.
\]

\[
\boxed{2}
\]

ANSWER 2: A

Problem 3:

The original paper is a \(4\)-inch by \(4\)-inch square.

First, it is folded vertically in half. The folded paper measures

\[
2\text{ inches by }4\text{ inches}.
\]

Then it is cut in half parallel to the fold. This creates:

- one large rectangle of dimensions \(2\) by \(4\),
- two small rectangles, each of dimensions \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10.
\]

The perimeter of the large rectangle is

\[
2(2+4)=12.
\]

So the desired ratio is

\[
\frac{10}{12}=\frac{5}{6}.
\]

\[
\boxed{\frac56}
\]

ANSWER 3: E

Problem 4:

We have

\[
\begin{array}{cccc}
 & T&W&O\\
+&T&W&O\\
\hline
F&O&U&R
\end{array}
\]

We are told that \(T=7\), and \(O\) is even.

Look at the hundreds column. Since \(7+7=14\), plus possibly a carry of \(0\) or \(1\), we get either \(14\) or \(15\).

Since the answer has thousands digit \(F\), we must have

\[
F=1.
\]

Also, the hundreds digit is \(O\). Thus \(O\) is either \(4\) or \(5\). Since \(O\) is even,

\[
O=4.
\]

So there was no carry from the tens column into the hundreds column.

Now look at the ones column:

\[
4+4=8.
\]

Thus

\[
R=8,
\]

with no carry into the tens column.

The tens column is therefore

\[
W+W=U,
\]

with no carry. Since the letters represent different digits, check the choices:

- \(W=0\) gives \(U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=3\) gives \(U=6\), which works.
- \(W=4\) is impossible because \(O=4\).

Thus

\[
W=3.
\]

Indeed,

\[
734+734=1468.
\]

\[
\boxed{3}
\]

ANSWER 4: D

Problem 5:

There are \(10\) possible tiles and \(6\) possible die results, for a total of

\[
10\cdot 6=60
\]

equally likely outcomes.

We count outcomes where the product is a perfect square.

A product is a square when the tile and die have the same squarefree part.

Check each tile:

\[
\begin{array}{c|c|c}
\text{Tile} & \text{Die results making a square} & \text{Number of results}\\
\hline
1 & 1,4 & 2\\
2 & 2 & 1\\
3 & 3 & 1\\
4 & 1,4 & 2\\
5 & 5 & 1\\
6 & 6 & 1\\
7 & \text{none} & 0\\
8 & 2 & 1\\
9 & 1,4 & 2\\
10 & \text{none} & 0
\end{array}
\]

For example:

\[
8\cdot2=16,
\]

which is a square, and

\[
9\cdot4=36,
\]

which is also a square.

The total number of favorable outcomes is

\[
2+1+1+2+1+1+0+1+2+0=11.
\]

Therefore the probability is

\[
\frac{11}{60}.
\]

\[
\boxed{\frac{11}{60}}
\]

ANSWER 5: C

Problem 6:

Let the original length be \(L\) and the original width be \(W\).

The original area is

\[
LW.
\]

Increasing the length by \(20\%\) multiplies it by

\[
1.20.
\]

Increasing the width by \(50\%\) multiplies it by

\[
1.50.
\]

So the new area is

\[
(1.20L)(1.50W)=1.80LW.
\]

The area is now \(180\%\) of the original area, so the increase is

\[
180\%-100\%=80\%.
\]

It is not \(20\%+50\%=70\%\), because both changes affect the area multiplicatively.

\[
\boxed{80\%}
\]

ANSWER 6: D

Problem 7:

The repeating sequence is

\[
ABCDEDCBAABCDEDCBA\cdots
\]

One full repeating block is

\[
ABCDEDCBA,
\]

which has \(9\) letters.

Now divide \(1992\) by \(9\):

\[
1992=9\cdot221+3.
\]

So the \(1992\)nd letter is the same as the 3rd letter in one block.

The third letter of

\[
ABCDEDCBA
\]

is \(C\).

\[
\boxed{C}
\]

ANSWER 7: C

Problem 8:

Each cube has:

- one protruding snap, and
- five receptacle holes.

For only receptacle holes to be showing, every protruding snap must be connected to another cube's hole.

Two cubes are not enough: if one cube’s protrusion snaps into the other, the second cube’s protrusion would still show.

Three cubes also cannot work. The protrusions would need to form a closed loop, but cubes joined face-to-face move in perpendicular grid directions, and three such moves cannot return to the starting location.

Four cubes can work by placing them in a \(2\) by \(2\) square arrangement. Each cube’s protrusion snaps into a hole on the next cube around the square:

\[
\text{Cube 1}\to\text{Cube 2}\to\text{Cube 3}\to\text{Cube 4}\to\text{Cube 1}.
\]

Then every protruding snap is hidden inside a connection, while the visible outside faces all contain receptacle holes.

Thus the smallest possible number is

\[
\boxed{4}.
\]

ANSWER 8: B

Problem 9:

Let \(f(n)\) be the number of ways Jo can climb \(n\) stairs.

To reach stair \(n\), Jo’s last step can be:

- \(1\) stair, after reaching \(n-1\),
- \(2\) stairs, after reaching \(n-2\),
- \(3\) stairs, after reaching \(n-3\).

Thus,

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

Start with:

\[
f(0)=1
\]

because there is one way to climb zero stairs: do nothing.

Then:

\[
f(1)=1
\]

since Jo must take