🔬 Model Lab

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around 

Problem 1:
We need TWO + TWO = FOUR with different digits for different letters, T=7, O even.

Write the column addition with carries:
```
  T W O
+ T W O
-------
F O U R
```
Let c1 be carry from units to tens, c2 tens to hundreds, c3 hundreds to thousands. Each is 0 or 1.

Units: 2O = R + 10 c1
Tens: 2W + c1 = U + 10 c2
Hundreds: 2T + c2 = O + 10 c3
Thousands: c3 = F

Since 2*TWO is a 4-digit number, c3 =1, so F=1.

Hundreds with T=7:
2*7 + c2 = O + 10*1
14 + c2 = O + 10
4 + c2 = O

c2 is 0 or 1, so O = 4 or 5. O is given even, so O = 4 and c2 = 0.

Units with O=4:
2*4 = 8 = R + 10 c1
So c1 must be 0 and R = 8.

Tens:
2W + c1 = U + 10 c2
2W + 0 = U + 0
So U = 2W and U < 10, hence 2W < 10, W < 5.

We already have T=7, F=1, O=4, R=8, c1=0, c2=0.
W must be distinct from 7,1,4,8 and <5.

Possibilities: 0,2,3.

- W=0 gives U=2W=0, duplicate W=U, not allowed since each letter is a different digit.
- W=2 gives U=4, duplicate U=O=4, not allowed.
- W=3 gives U=6, all digits 7,3,4,1,6,8 distinct.

Check: 734+734=1468, which is FOUR. Works.

So only W=3 is possible.
ANSWER 1: D

Problem 2:
Let N be the total number of people. 2/5 of them wear gloves and 3/4 wear hats, so N must allow integers.

N must be divisible by 5 and by 4, hence by 20. Write N=20k for integer k>=1.

Gloves: (2/5)N = 8k
Hats: (3/4)N = 15k

Let B be both. By inclusion-exclusion:
|Gloves or Hats| = |Gloves| + |Hats| - |Both| <= N

So |Both| >= |Gloves|+|Hats|-N = 8k+15k-20k = 3k.

The overlap is at least 3k, attained when everyone wears at least one of the two. The smallest room has k=1, N=20, giving overlap at least 3.

Hence minimum wearing both is 3.
ANSWER 2: A

Problem 3:
Car uses 1 gallon per 35 miles. Tank holds 14 gallons.

Start full: 14 gal.
Drive 350 miles: uses 350/35 = 10 gal.
Remaining: 14-10 = 4 gal.
Buy 8 gal: 4+8 = 12 gal.

Drive further x miles to destination, using x/35 gal, arriving half-full.
Half-full = 14/2 = 7 gal.

So 12 - x/35 = 7
x/35 = 5
x = 175

Total miles = 350 + 175 = 525.
ANSWER 3: A

Problem 4:
Let 1st tank have a guppies.
2nd = a+1
3rd = (a+1)+2 = a+3
4th = (a+3)+3 = a+6

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

4th tank = a+6 = 26.
ANSWER 4: E

Problem 5:
Only 6 can race at a time, and each race eliminates the 5 losers. The winner stays.

To get from 216 sprinters to 1 champion, 215 sprinters must be eliminated.

Each race eliminates exactly 5 sprinters. So number of races needed = 215/5 = 43.

Check by staging: 216/6=36 races give 36 winners; 36/6=6 races give 6 winners; 1 final race gives champion. Total 36+6+1=43.

Choices 36 forgets later rounds; 42 is off by one.
ANSWER 5: C

Problem 6:
Mileage travelled = 57,060 - 56,200 = 860 miles.

The driver "filled the tank" at the start, again mid-trip, and again at the end. That means the tank is full after each fill. The 6 gallons at the start brought it to full before the trip; they do not count as fuel burned on the trip.

Leg 1: 56,560 - 56,200 = 360 miles. Fuel burned to be replaced by the 12-gallon fill = 12 gal.
Leg 2: 57,060 - 56,560 = 500 miles. Fuel burned to be replaced by the 20-gallon fill = 20 gal.

Fuel burned on trip = 12+20 = 32 gal.

Average mpg = 860 / 32 = 26.875 ≈ 26.9 to nearest tenth.

If you mistakenly include all 6+12+20=38 gal you get 22.6, but then the intermediate odometer reading would be useless, and the initial 6 gal was not burned on this trip.
ANSWER 6: D

Problem 7:
Let a = jumps of +5 to the right, b = jumps of -3 to the left. Total jumps n=a+b.

Net: 5a - 3b = 2023, a,b >=0 integers. Minimize n.

5a = 2023+3b must be divisible by 5.
2023 ≡ 3 mod 5, so 3+3b ≡ 0 mod 5, i.e. 3b ≡ 2 mod 5.
Since 3*2=6≡1 mod5, multiply by 2: b ≡ 4 mod5.

Write b = 4+5t, t>=0 integer.
Then a = (2023+3b)/5 = (2023+12+15t)/5 = (2035+15t)/5 = 407+3t.

Total n = a+b = 407+3t+4+5t = 411+8t.

This is smallest for smallest t, t=0, giving b=4, a=407, n=411.
Check: 407*5 - 4*3 = 2035-12=2023.
ANSWER 7: D

Problem 8:
Sides are 6.5, 10, s integer. Triangle inequality: sum of any two sides > third side.

In particular the two shorter sides must sum to more than the longest.

If s is small, s+6.5 > 10 is required:
s > 3.5. Since s is whole, s >= 4.

Check s=4: 4+6.5=10.5>10, 4+10>6.5, 6.5+10>4, works.
s=3 fails: 3+6.5=9.5<10.

Also need 6.5+10 > s, so s<16.5, satisfied.

Smallest possible is 4.
ANSWER 8: B

Problem 9:
360 = 36*10 = 2^3 * 3^2 * 5^1.

For 360x to be a square, all prime exponents must be even.
2-exponent 3 is odd -> need one more 2.
3-exponent 2 is even -> need none.
5-exponent 1 is odd -> need one more 5.
Smallest x = 2*5 = 10. Indeed 360*10=3600=60^2.

For 360y to be a cube, all exponents must be multiples of 3.
2-exponent 3 already multiple of 3 -> need none.
3-exponent 2 -> need one more to get 3.
5-exponent 1 -> need two more to get 3.
Smallest y = 3^1 * 5^2 = 3*25 = 75. Indeed 360*75=27000=30^3.

Sum = 10+75 = 85.
ANSWER 9: B

Problem 10:
To surround the patio needs bushes along the circumference, spaced ~1 foot apart.

Circumference = 2πr ≈ 2*3.14*12 ≈ 75.4 feet.

At 1 foot spacing, need about 75 bushes.

450 would be about the area, πr^2≈452, not the perimeter.
ANSWER 10: D

Problem 1:
**Restatement:** 
We are given the cryptarithm:
$$\begin{array}{cccc} & \text{T} & \text{W} & \text{O} \\ + & \text{T} & \text{W} & \text{O} \\ \hline \text{F} & \text{O} & \text{U} & \text{R} \end{array}$$
Each letter represents a unique digit from $0$ to $9$. We are given $T = 7$, and $O$ is an even number. We must determine the only possible value for $W$.

**Step-by-step reasoning:**
1. Substituting $T = 7$:
   $$7\text{WO} + 7\text{WO} = \text{FOUR}$$
2. The sum of two 3-digit numbers starting with $7$ is at least $700 + 700 = 1400$ and at most $799 + 799 = 1598$. 
   Therefore, the thousands digit must be $F = 1$.
3. Since $F = 1$, the hundreds digit $O$ can only be $4$ or $5$ (from $7 + 7 = 14$ or $7 + 7 + 1 = 15$). 
   We are given that $O$ is an even digit, so $O = 4$.
4. Since $O = 4$, looking at the ones column:
   $$O + O = 4 + 4 = 8$$
   This produces no carry into the tens column, and gives $R = 8$.
5. Now looking at the hundreds column:
   $$T + T + (\text{carry from } W + W) = 7 + 7 + \text{carry} = 14$$
   This means there is **no carry** from the tens column ($W + W$) to the hundreds column.
   Therefore, $2W < 10$, which implies $W \in \{0, 1, 2, 3, 4\}$.
6. The digits already assigned to other letters are:
   $$F = 1, \quad O = 4, \quad T = 7, \quad R = 8$$
   - If $W = 0$, then $U = 2W = 0$, so $U = W$, which violates the rule that letters represent distinct digits.
   - If $W = 1$, then $W = F = 1$, which is not allowed.
   - If $W = 2$, then $U = 2W = 4$, which means $U = O = 4$, not allowed.
   - If $W = 4$, then $W = O = 4$, not allowed.
   - If $W = 3$, then $U = 2W = 6$. The digits used are $T=7, W=3, O=4, F=1, U=6, R=8$, all of which are distinct.

Thus, the only possible value for $W$ is $3$.

ANSWER 1: D

---

Problem 2:
**Restatement:** 
In a room, $\frac{2}{5}$ of the people wear gloves and $\frac{3}{4}$ wear hats. What is the minimum number of people in the room wearing both?

**Step-by-step reasoning:**
1. Let $N$ be the total number of people in the room. The number of people wearing gloves, $\frac{2}{5}N$, and hats, $\frac{3}{4}N$, must both be integers. Thus, $N$ must be a multiple of $\text{lcm}(5, 4) = 20$.
2. To find the minimum number of people who wear both, we should minimize the total population $N$, so we test the smallest positive integer $N = 20$.
3. With $N = 20$:
   - People wearing gloves: $|G| = \frac{2}{5} \times 20 = 8$.
   - People wearing hats: $|H| = \frac{3}{4} \times 20 = 15$.
4. By the Principle of Inclusion-Exclusion:
   $$|G \cap H| = |G| + |H| - |G \cup H|$$
   Since $|G \cup H| \le N = 20$, we have:
   $$|G \cap H| \ge 8 + 15 - 20 = 3$$
5. This minimum of $3$ people wearing both is achievable (e.g., $5$ wear only gloves, $12$ wear only hats, and $3$ wear both, totaling $5 + 12 + 3 = 20$ people).
- Other choices like $5, 8, 15, 20$ represent larger overlaps or total group sizes, but not the minimum possible overlap.

ANSWER 2: A

---

Problem 3:
**Restatement:** 
Karl's car gets 35 miles per gallon and has a 14-gallon tank. Karl starts with a full tank, drives 350 miles, buys 8 gallons of gas, and drives until the tank is half full. How many total miles did Karl drive that day?

**Step-by-step reasoning:**
1. Karl starts with a full tank: $14$ gallons.
2. In the first leg of the trip, he drives $350$ miles:
   $$\text{Gas used} = \frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
3. Gas remaining in the tank:
   $$14 - 10 = 4\text{ gallons}$$
4. He buys $8$ gallons of gas, so the tank now has:
   $$4 + 8 = 12\text{ gallons}$$
5. At the destination, the tank is half full:
   $$\frac{14}{2} = 7\text{ gallons}$$
6. In the second leg of the trip, the gas used is:
   $$12 - 7 = 5\text{ gallons}$$
7. The distance driven in the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$
8. Total distance driven:
   $$350 + 175 = 525\text{ miles}$$

ANSWER 3: A

---

Problem 4:
**Restatement:** 
Rohan has $90$ guppies distributed among $4$ fish tanks such that:
- Tank 2 has 1 more guppy than Tank 1.
- Tank 3 has 2 more guppies than Tank 2.
- Tank 4 has 3 more guppies than Tank 3.
How many guppies are in Tank 4?

**Step-by-step reasoning:**
1. Let the number of guppies in Tank 1 be $x$.
2. Express the number of guppies in each tank in terms of $x$:
   - $\text{Tank 1} = x$
   - $\text{Tank 2} = x + 1$
   - $\text{Tank 3} = (x + 1) + 2 = x + 3$
   - $\text{Tank 4} = (x + 3) + 3 = x + 6$
3. Sum the guppies in all four tanks:
   $$x + (x + 1) + (x + 3) + (x + 6) = 90$$
   $$4x + 10 = 90$$
   $$4x = 80 \implies x = 20$$
4. Find the number of guppies in the 4th tank:
   $$\text{Tank 4} = x + 6 = 20 + 6 = 26$$
- The other choices correspond to: Tank 1 ($20$), Tank 2 ($21$), Tank 3 ($23$), and an arithmetic error ($24$).

ANSWER 4: E

---

Problem 5:
**Restatement:** 
216 sprinters compete in a 100-meter dash. The track has 6 lanes. In each race of 6 sprinters, 5 non-winners are eliminated and only 1 winner advances. How many races are needed to determine the single champion?

**Step-by-step reasoning:**
1. **Method 1 (Elimination approach):**
   - There are $216$ sprinters and only $1$ champion at the end.
   - Therefore, exactly $216 - 1 = 215$ sprinters must be eliminated.
   - Each race eliminates exactly $5$ sprinters.
   - Number of races required:
     $$\frac{215}{5} = 43\text{ races}$$

2. **Method 2 (Round-by-round approach):**
   - Round 1: $\frac{216}{6} = 36$ races, leaving $36$ winners.
   - Round 2: $\frac{36}{6} = 6$ races, leaving $6$ winners.
   - Round 3: $\frac{6}{6} = 1$ race, leaving $1$ winner.
   - Total races: $36 + 6 + 1 = 43$ races.

ANSWER 5: C

---

Problem 6:
**Restatement:** 
At the start, the odometer read $56{,}200$ miles and the driver filled the tank with $6$ gallons. During the trip, he filled the tank with $12$ gallons at odometer reading $56{,}560$. At the end, he filled the tank with $20$ gallons at odometer reading $57{,}060$. What was the average miles-per-gallon

# Problem 1:

**Setup:** TWO + TWO = FOUR, with T = 7, all letters distinct digits, and O is even.

**Analysis:** The sum is 2 × TWO = 2 × (700 + 10W + O) = 1400 + 20W + 2O.

Since 20W + 2O is at most 198, the sum lies between 1400 and 1598. So the thousands digit is F = 1, and the **hundreds digit** of the sum is either 4 or 5.

But in FOUR, the hundreds digit is the letter **O**. Since O is even, O cannot be 5, so **O = 4**.

**Finding W:** With O = 4, the sum is 1400 + 20W + 8 = 1408 + 20W. For the hundreds digit to be 4 (i.e., the sum stays under 1500), we need 20W ≤ 90, so W ≤ 4. Now test W = 0, 2, 3 (W ≠ 4 = O, W ≠ 7, W ≠ 1 = F):

- **W = 0:** TWO = 704, sum = 1408. Tens digit U = 0 = W. ✗ (letters must differ)
- **W = 2:** TWO = 724, sum = 1448. Tens digit U = 4 = O. ✗
- **W = 3:** TWO = 734, sum = 734 × 2 = 1468. Check: F=1, O=4, U=6, R=8 — all distinct from T=7 and W=3. ✓

So the only possible value is W = 3. (This also rules out tempting guesses like W = 2, which fail the distinctness rule.)

ANSWER 1: D

# Problem 2:

**Setup:** Let the total be N people. Glove-wearers: (2/5)N; hat-wearers: (3/4)N.

**Key idea (Inclusion-Exclusion):**
Both ≥ Glove + Hat − Total = (2/5)N + (3/4)N − N = (8/20 + 15/20 − 20/20)N = (3/20)N.

**Minimum:** The number of people must be divisible by both 5 and 4, so the smallest possible total is N = 20. Then at least (3/20)(20) = 3 people wear both. (Check: 8 wear gloves, 15 wear hats; 8 + 15 = 23, so at least 3 of the 20 wear both — achievable.)

ANSWER 2: A

# Problem 3:

**Setup:** Karl starts with a full tank of 14 gallons.

**Step 1:** Driving 350 miles at 35 miles/gallon uses 350 ÷ 35 = 10 gallons, leaving 14 − 10 = 4 gallons.

**Step 2:** He buys 8 gallons, so the tank now has 4 + 8 = 12 gallons.

**Step 3:** He arrives with a half-full tank: 14 ÷ 2 = 7 gallons. So on the last leg he used 12 − 7 = 5 gallons, covering 5 × 35 = 175 miles.

**Total:** 350 + 175 = 525 miles.

(Tempting wrong answer 560 would come from forgetting the tank wasn't full after the purchase.)

ANSWER 3: A

# Problem 4:

**Setup:** Let the first tank have g guppies. Then the tanks hold:
g, g + 1, g + 2, g + 3.

**Equation:** g + (g+1) + (g+2) + (g+3) = 90 → 4g + 6 = 90 → 4g = 84 → g = 21.

**Fourth tank:** g + 3 = 21 + 3 = 24 guppies. (Check: 21 + 22 + 23 + 24 = 90. ✓)

ANSWER 4: D

# Problem 5:

**Key insight:** Each race eliminates exactly 5 sprinters (the five non-winners), and the winner continues.

**Setup:** Start with 216 sprinters; end with 1 champion. So the number of sprinters eliminated is 216 − 1 = 215.

**Computation:** Each race eliminates 5 sprinters, so the number of races needed is 215 ÷ 5 = 43.

(A tempting wrong answer is 36 = 216 ÷ 6, but that ignores that winners keep racing. The answer must be slightly more than 216/6 because winners are re-used.)

ANSWER 5: C

# Problem 6:

**Setup:** The total gasoline used during the trip is the sum of all fill-ups that "replaced" gas burned: 6 + 12 + 20 = 38 gallons.

**Total distance:** 57,060 − 56,200 = 860 miles.

**Average:** 860 ÷ 38 ≈ 22.631... ≈ 22.6 miles per gallon.

(A common mistake is to divide 860 by 32 (12+20) or use only part of the gas — those don't match the total fuel consumed.)

ANSWER 6: B

# Problem 7:

**Setup:** Suppose Greta makes a right-jumps and b left-jumps. Each right jump gains +5, each left jump −3:
5a − 3b = 2023, and we minimize a + b.

**Modular condition:** Reduce mod 3: 5a ≡ 2023 (mod 3) → 2a ≡ 1 (mod 3) → a ≡ 2 (mod 3).

**Minimizing jumps:** Since right jumps move 5 pads versus only 3 for left jumps, use as many right jumps as possible. Also 5a ≥ 2023 forces a ≥ 405 (since 5 × 404 = 2020 < 2023).

Checking a ≡ 2 (mod 3): a = 405 gives 5 × 405 = 2025, needing 3b = 2 — not an integer. Next value a = 407: 5 × 407 = 2035, so 3b = 12 → b = 4.

**Total:** a + b = 407 + 4 = 411 jumps.

(Check: 5 × 407 − 3 × 4 = 2035 − 12 = 2023. ✓)

ANSWER 7: D

# Problem 8:

**Triangle inequality:** The sum of any two sides must exceed the third. With sides 6.5, 10, and s:

- s + 6.5 > 10 → s > 3.5
- s + 10 > 6.5 → always true for positive s
- 6.5 + 10 > s → s < 16.5

**Smallest whole number:** s must be greater than 3.5, so the smallest whole number is s = 4.

(Check: 4 + 6.5 = 10.5 > 10. ✓ — while s = 3 fails since 3 + 6.5 = 9.5 < 10.)

ANSWER 8: B

# Problem 9:

**Factor 360:** 360 = 2³ · 3² · 5.

**Finding x (smallest making 360x a perfect square):** A square needs all prime exponents even. Currently: 2³ (odd), 3² (even), 5¹ (odd). Multiply by x = 2 · 5 = 10:
360 × 10 = 2⁴ · 3² · 5² = (2² · 3 · 5)² = 60². ✓ So **x = 10**.

**Finding y (smallest making 360y a perfect cube):** A cube needs all prime exponents divisible by 3. Currently: 2³ (OK), 3² (need one more 3), 5¹ (need two more 5s). Multiply by y = 3 · 5² = 3 · 25 = 75:
360 × 75 = 2⁹ · 3³ · 5³ = (2³ · 3 · 5)³. ✓ So **y = 75**.

**Sum:** x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:

**Setup:** The bushes surround a circle of radius 12 feet, spaced about 1 foot apart. The number of bushes needed is approximately the circumference (in feet).

**Circumference:** C = 2πr = 2 · π · 12 = 24π ≈ 24 × 3.14159 ≈ 75.4 feet.

**Conclusion:** About 75 bushes are needed.

(Ruling out others: 450 would come from wrongly using an area formula; 12 from using just the radius; 38 from using πr² ... none match the circumference ≈ 75.)

ANSWER 10: D

Problem 1:
We need to solve

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

with \(T=7\), \(O\) even, and all letters representing different digits.

First note that \(TWO\) is between \(700\) and \(799\). Doubling it gives a number between \(1400\) and \(1598\), so the thousands digit must be

\[
F=1.
\]

Now work column by column. Let \(c_1\) be the carry from the ones column and \(c_2\) the carry from the tens column.

Ones column:

\[
2O=R+10c_1.
\]

Tens column:

\[
2W+c_1=U+10c_2.
\]

Hundreds column:

\[
2T+c_2=O+10F.
\]

Since \(T=7\) and \(F=1\),

\[
14+c_2=O+10,
\]

so

\[
O=4+c_2.
\]

The tens carry \(c_2\) can only be \(0\) or \(1\), because \(2W+c_1\le 19\).

- If \(c_2=0\), then \(O=4\), which is even.
- If \(c_2=1\), then \(O=5\), which is odd, so not allowed.

Thus \(c_2=0\) and \(O=4\).

From the ones column,

\[
2(4)=8,
\]

so \(R=8\) and \(c_1=0\).

From the tens column,

\[
2W=U.
\]

Now check the answer choices for \(W\):

- \(W=0\Rightarrow U=0\), but then \(W\) and \(U\) are the same digit.
- \(W=1\), but \(F=1\), so \(W\) and \(F\) are the same digit.
- \(W=2\Rightarrow U=4\), but \(O=4\).
- \(W=3\Rightarrow U=6\), and all digits \(T=7,W=3,O=4,F=1,U=6,R=8\) are different.
- \(W=4\), but \(O=4\).

So the only possible value is \(W=3\).

ANSWER 1: D

Problem 2:
Let the total number of people be \(N\). Since \(\frac25\) of the people wear gloves and \(\frac34\) wear hats, \(N\) must be a multiple of both \(5\) and \(4\), so \(N\) must be a multiple of \(20\).

The number wearing both is at least

\[
\frac25N+\frac34N-N
=
\left(\frac{8}{20}+\frac{15}{20}-\frac{20}{20}\right)N
=
\frac{3}{20}N.
\]

The smallest positive multiple of \(20\) is \(N=20\). Then the minimum number wearing both is

\[
\frac{3}{20}\cdot 20=3.
\]

This is attainable: with \(20\) people, \(8\) wear gloves, \(15\) wear hats, and exactly \(3\) can wear both.

ANSWER 2: A

Problem 3:
Karl’s car uses \(1\) gallon per \(35\) miles, so it uses

\[
\frac{350}{35}=10
\]

gallons for the first \(350\) miles.

He started with a full \(14\)-gallon tank, so after driving \(350\) miles he had

\[
14-10=4
\]

gallons left.

He then bought \(8\) gallons, so he had

\[
4+8=12
\]

gallons.

When he arrived, the tank was half full. Since a full tank holds \(14\) gallons, half full means

\[
7
\]

gallons.

So after buying the gas, he used

\[
12-7=5
\]

gallons, which corresponds to

\[
5\cdot 35=175
\]

miles.

Total miles driven:

\[
350+175=525.
\]

ANSWER 3: A

Problem 4:
Let the number of guppies in the 1st tank be \(x\).

Then:

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

Total guppies:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine terms:

\[
4x+10=90.
\]

So

\[
4x=80,
\]

and

\[
x=20.
\]

The 4th tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:
Each race has \(6\) sprinters. At the end, \(5\) are eliminated and \(1\) continues.

Thus each race eliminates exactly \(5\) sprinters.

There are \(216\) sprinters at the start, and we need only \(1\) champion left. Therefore the number of eliminations needed is

\[
216-1=215.
\]

Since each race eliminates \(5\), the number of races is

\[
\frac{215}{5}=43.
\]

We can also check the bracket:

\[
216\to 36 \to 6 \to 1,
\]

using \(36+6+1=43\) races.

ANSWER 5: C

Problem 6:
The odometer readings go from \(56,200\) to \(57,060\), so the trip distance is

\[
57,060-56,200=860
\]

miles.

At the start of the trip, the tank is full. When the driver next fills the tank and adds \(12\) gallons, the car must have used those \(12\) gallons during the first part of the trip.

Then from that point until the end, he fills the tank with \(20\) gallons, so the car used \(20\) gallons during the second part.

Thus the total gasoline used during the trip is

\[
12+20=32
\]

gallons.

Average miles per gallon:

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth, this is

\[
26.9.
\]

The tempting but incorrect choice \(22.6\) comes from dividing by \(6+12+20=38\) gallons, but the initial \(6\) gallons are not part of the net gas used during the trip.

ANSWER 6: D

Problem 7:
Let \(a\) be the number of jumps \(5\) pads to the right, and let \(b\) be the number of jumps \(3\) pads to the left.

The net movement is

\[
5a-3b=2023.
\]

We want to minimize the total number of jumps:

\[
a+b.
\]

Solve for \(b\):

\[
3b=5a-2023,
\]

so

\[
b=\frac{5a-2023}{3}.
\]

For \(b\) to be an integer, \(5a-2023\) must be divisible by \(3\). Since \(5\equiv 2\pmod 3\) and \(2023\equiv 1\pmod 3\),

\[
2a-1\equiv 0\pmod 3,
\]

so

\[
2a\equiv 1\pmod 3.
\]

Multiplying by \(2\), which is its own inverse modulo \(3\),

\[
a\equiv 2\pmod 3.
\]

Also, we need

\[
5a\ge 2023,
\]

so

\[
a\ge \frac{2023}{5}=404.6,
\]

meaning

\[
a\ge 405.
\]

The smallest integer at least \(405\) that is congruent to \(2\pmod 3\) is

\[
a=407.
\]

Then

\[
b=\frac{5(407)-2023}{3}
=
\frac{2035-2023}{3}
=
\frac{12}{3}=4.
\]

Total jumps:

\[
a+b=407+4=411.
\]

Larger valid values of \(a\) would only increase the total number of jumps.

ANSWER 7: D

Problem 8:
The triangle sides are \(6.5\), \(10\), and \(s\), where \(s\) is a whole number.

The triangle inequality says the sum of any two sides must be greater than the third side. The most restrictive condition is

\[
s+6.5>10.
\]

So

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

Check:

\[
4+6.5=10.5>10,
\]

so a triangle with sides \(4,6.5,10\) is possible. The choice \(3\) fails because \(3+6.5=9.5<10\).

ANSWER 8: B

Problem 9:
Factor \(360\):

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a perfect square, all prime exponents must be even.

Current exponents:

- \(2\): exponent \(3\), needs one more factor of \(2\)
- \(3\): exponent \(2\), already even
- \(5\): exponent \(1\), needs one more factor of \(5\)

So the smallest \(x\) is

\[
x=2\cdot 5=10.
\]

For \(360y\) to be a perfect cube, all prime exponents must 

# Problem 1:
We need 2·(TWO) = FOUR with T = 7, O even, all letters different digits.

Set up the columns of 7W O + 7W O:
- **Units:** O + O = R (+ carry c₁)
- **Tens:** W + W + c₁ = U (+ carry c₂)
- **Hundreds:** 7 + 7 + c₂ = O (+ carry c₃)
- **Thousands:** c₃ = F

From the hundreds column: 14 + c₂ is either 14 or 15, so the hundreds digit O = 4 or 5, and F = 1.
Since O must be even, **O = 4**, which forces c₂ = 0.

Units: 4 + 4 = 8, so R = 8 with c₁ = 0.
Tens: 2W = U with no carry (c₂ = 0), so 2W < 10.

Now test W (must differ from T=7, F=1, O=4, R=8):
- W = 0 → U = 0, same as W. ✗
- W = 1 → equals F. ✗
- W = 2 → U = 4, same as O. ✗
- W = 3 → U = 6: 734 + 734 = 1468 ✓ (digits 7,3,4,1,6,8 all different)
- W = 4 → equals O (and U = 8 = R). ✗

Only W = 3 works.

ANSWER 1: D

# Problem 2:
The number of people must make both 2/5 and 3/4 whole numbers, so it's a multiple of 20; the smallest room size is 20 people.

- Gloves: (2/5)(20) = 8
- Hats: (3/4)(20) = 15

Minimum overlap = 8 + 15 − 20 = 3 (we can't have 0 overlap since 8 + 15 = 23 > 20).

(Larger rooms, e.g. 40 people, give overlap ≥ 16+30−40 = 6, so 3 is truly minimal.)

ANSWER 2: A

# Problem 3:
Start: 14 gallons.
- 350 miles uses 350/35 = 10 gallons → 4 gallons left.
- Buys 8 gallons → 12 gallons.
- Arrives with half a tank = 7 gallons, so he used 12 − 7 = 5 gallons.
- Those 5 gallons = 5 × 35 = 175 miles.

Total = 350 + 175 = 525 miles.

ANSWER 3: A

# Problem 4:
Let the 1st tank have x guppies. Then tanks are x, x+1, x+3, x+6.

x + (x+1) + (x+3) + (x+6) = 90 → 4x + 10 = 90 → x = 20.

Tanks: 20, 21, 23, 26. The 4th tank has 26.

(Note 20, 21, 23 are the tempting distractors — they're the other tanks.)

ANSWER 4: E

# Problem 5:
Each race eliminates exactly 5 sprinters (the winner survives). To crown one champion, 216 − 1 = 215 sprinters must be eliminated.

215 ÷ 5 = 43 races.

ANSWER 5: C

# Problem 6:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only filled the tank *before* the trip; the gas actually consumed during the trip is what was replaced afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ 26.9 mpg.

(Using 38 gallons gives 22.6 — that's the trap answer B.)

ANSWER 6: D

# Problem 7:
Let a = jumps right (+5), b = jumps left (−3), total jumps n = a + b.

5a − 3b = 2023, with b = n − a:
5a − 3(n − a) = 8a − 3n = 2023 → 8a = 2023 + 3n.

Since a ≤ n: 2023 + 3n ≤ 8n → n ≥ 404.6, so n ≥ 405.

Also 8a = 2023 + 3n requires 2023 + 3n ≡ 0 (mod 8). Since 2023 ≡ 7 (mod 8): 7 + 3n ≡ 0 → 3n ≡ 1 (mod 8) → n ≡ 3 (mod 8).

Smallest n ≥ 405 with n ≡ 3 (mod 8) is n = 411 (then a = (2023 + 1233)/8 = 407, b = 4; check 5·407 − 3·4 = 2035 − 12 = 2023 ✓).

ANSWER 7: D

# Problem 8:
Triangle inequality: s + 6.5 > 10 → s > 3.5.

Since s is a whole number, the smallest is s = 4 (check: 4 + 6.5 = 10.5 > 10 ✓, 4 + 10 > 6.5 ✓, 6.5 + 10 > 4 ✓).

s = 3 fails since 3 + 6.5 = 9.5 < 10.

ANSWER 8: B

# Problem 9:
360 = 2³ · 3² · 5.

**Square:** exponents must all be even. Need one more 2 (3→4) and one more 5 (1→2): x = 2·5 = 10. Check 360·10 = 3600 = 60². ✓

**Cube:** exponents must be multiples of 3. 2³ is fine; need one more 3 (2→3) and two more 5s (1→3): y = 3·5² = 75. Check 360·75 = 27,000 = 30³. ✓

x + y = 10 + 75 = 85.

ANSWER 9: B

# Problem 10:
The bushes go around the circle's edge: circumference = 2πr = 2π(12) = 24π ≈ 75.4 feet.

With bushes 1 foot apart, about 75 bushes are needed.

(48 comes from mistakenly using 4r or diameter-based reasoning; 450 is the area 144π ≈ 452.)

ANSWER 10: D

Problem 1:

We need solve the addition

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

where all letters represent different digits, \(T=7\), and \(O\) is even.

In the hundreds column, we add \(7+7\), plus possibly a carry from the tens column.

\[
7+7=14
\]

So the thousands digit must be \(F=1\). The hundreds digit is either \(4\) or \(5\), depending on whether there is a carry from the tens column.

But \(O\) is even, so

\[
O=4.
\]

Now look at the ones column:

\[
O+O=4+4=8.
\]

Thus \(R=8\), with no carry into the tens column.

The tens column is therefore

\[
W+W=U.
\]

There cannot be a carry, since a carry would have made \(O=5\). Thus \(2W<10\), so \(W\) can only be \(0,1,2,3,\) or \(4\).

Check the possibilities:

- \(W=0\) gives \(U=0\), but letters must represent different digits.
- \(W=1\) is impossible because \(F=1\).
- \(W=2\) gives \(U=4\), but \(O=4\).
- \(W=4\) is impossible because \(O=4\).
- \(W=3\) gives \(U=6\), which is different from \(1,4,7,8\).

Therefore,

\[
W=3.
\]

ANSWER 1: D

Problem 2:

Let the total number of people be \(N\).

Since \(\frac25\) wear gloves and \(\frac34\) wear hats, \(N\) must be divisible by both \(5\) and \(4\). The smallest possible total is therefore

\[
\operatorname{lcm}(5,4)=20.
\]

With 20 people:

\[
\frac25(20)=8
\]

people wear gloves, and

\[
\frac34(20)=15
\]

people wear hats.

If nobody wore both, there would need to be

\[
8+15=23
\]

different people, but there are only 20 people total. Thus at least

\[
23-20=3
\]

people must wear both hats and gloves.

This is possible: 3 can wear both, 5 can wear only gloves, and 12 can wear only hats.

So the minimum is \(3\). Choices such as 5, 8, 15, and 20 are possible numbers in other arrangements, but they are not the minimum.

ANSWER 2: A

Problem 3:

Karl starts with 14 gallons.

His car travels 35 miles per gallon. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has

\[
14-10=4
\]

gallons left.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, his 14-gallon tank is half full, so he has

\[
\frac{14}{2}=7
\]

gallons remaining.

Therefore, during the second part of the trip he used

\[
12-7=5
\]

gallons. That distance was

\[
5\cdot 35=175
\]

miles.

His total distance was

\[
350+175=525
\]

miles.

ANSWER 3: A

Problem 4:

Let the number of guppies in the first tank be \(x\).

Then the tanks contain:

\[
\begin{aligned}
\text{1st tank} &=x,\\
\text{2nd tank} &=x+1,\\
\text{3rd tank} &=x+1+2=x+3,\\
\text{4th tank} &=x+3+3=x+6.
\end{aligned}
\]

Their total is 90:

\[
x+(x+1)+(x+3)+(x+6)=90.
\]

Combine like terms:

\[
4x+10=90.
\]

\[
4x=80
\]

\[
x=20.
\]

Thus the fourth tank has

\[
x+6=20+6=26
\]

guppies.

ANSWER 4: E

Problem 5:

Each race has 6 sprinters, and 5 are eliminated. Therefore, every race reduces the number of remaining sprinters by 5.

To reduce from 216 sprinters to 1 champion, we must eliminate

\[
216-1=215
\]

sprinters.

Since each race eliminates 5 sprinters, the number of races needed is

\[
\frac{215}{5}=43.
\]

Another way to see this is:

- First round: \(216\div 6=36\) races, leaving 36 winners.
- Next round: \(36\div 6=6\) races, leaving 6 winners.
- Final: 1 race among those 6 winners.

So the total is

\[
36+6+1=43.
\]

The choice 36 counts only the first round, not the later races among the winners.

ANSWER 5: C

Problem 6:

The total distance driven was

\[
57,\!060-56,\!200=860
\]

miles.

At the beginning, the tank was filled. At the end, it was filled again, so the tank had the same amount of gasoline at the beginning and end of the trip.

Thus, the gas used during the trip equals the gasoline bought after starting:

\[
12+20=32
\]

gallons.

The initial 6 gallons filled before the trip should not be added, because that established the full starting tank.

Therefore, average miles per gallon was

\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,

\[
26.875\approx 26.9.
\]

ANSWER 6: D

Problem 7:

Suppose Greta makes \(r\) jumps to the right and \(l\) jumps to the left.

Each right jump moves her 5 pads right, and each left jump moves her 3 pads left. So

\[
5r-3l=2023.
\]

The total number of jumps is

\[
r+l.
\]

Let this total be \(n\). Since \(r=n-l\),

\[
5(n-l)-3l=2023.
\]

\[
5n-8l=2023.
\]

Thus

\[
5n=2023+8l.
\]

For the right side to be divisible by 5, we need

\[
5n\equiv 2023 \pmod 8.
\]

Since \(2023\equiv 7\pmod 8\), we need

\[
5n\equiv 7\pmod 8.
\]

Because \(5\cdot 5=25\equiv 1\pmod 8\), multiply both sides by 5:

\[
n\equiv 35\equiv 3\pmod 8.
\]

Also, even if every jump were rightward, Greta would need at least

\[
\left\lceil\frac{2023}{5}\right\rceil=405
\]

jumps.

The first number at least 405 that is congruent to 3 modulo 8 is

\[
411.
\]

Check that it works:

\[
5(411)-2023=2055-2023=32,
\]

and

\[
32=8(4).
\]

So \(l=4\) left jumps, and then

\[
r=411-4=407
\]

right jumps. The net movement is

\[
407(5)-4(3)=2035-12=2023.
\]

ANSWER 7: D

Problem 8:

For side lengths \(6.5\), \(10\), and \(s\) to form a triangle, the sum of the two shorter sides must be greater than the longest side.

In particular,

\[
6.5+s>10.
\]

Subtract 6.5:

\[
s>3.5.
\]

Since \(s\) is a whole number, the smallest possible value is

\[
s=4.
\]

The choice \(3\) fails because

\[
6.5+3=9.5<10,
\]

so those lengths cannot form a triangle.

ANSWER 8: B

Problem 9:

First factor 360:

\[
360=2^3\cdot 3^2\cdot 5.
\]

For \(360x\) to be a square, every prime exponent must be even.

Currently the exponents are \(3,2,1\). We need one more factor of 2 and one more factor of 5:

\[
x=2\cdot 5=10.
\]

Then

\[
360x=2^4\cdot 3^2\cdot 5^2,
\]

which is a square.

For \(360y\) to be a cube, every prime exponent must be a multiple of 3.

- \(2^3\) is already fine.
- \(3^2\) needs one more factor of 3.
- \(5^1\) needs two more factors of 5.

Thus

\[
y=3\cdot 5^2=75.
\]

Therefore,

\[
x+y=10+75=85.
\]

ANSWER 9: B

Problem 10:

The rose bushes go around