🔬 Model Lab

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu

Problem 1:
We need 9 coins with at least one penny, nickel, dime, quarter, total 102 cents.
Let counts be $p,n,d,q \ge 1$ with $p+n+d+q=9$.

Value in cents:
$$p+5n+10d+25q=102.$$

Substitute $p=9-n-d-q$:
$$(9-n-d-q)+5n+10d+25q=102$$
$$9+4n+9d+24q=102$$
$$4n+9d+24q=93.$$

$q$ cannot be large. Since $n\ge1,d\ge1$:
$24q \le 93-4-9=80$, so $q=1,2,3$.

- $q=3$: $4n+9d=21$. Try $d=1$: $4n=12$, $n=3$. Then $p=9-3-1-3=2\ge1$. This works.
$d\ge2$ gives $4n\le3$, impossible.

- $q=2$: $4n+9d=45$. Mod 4, $d\equiv1\pmod4$. $d=1$ gives $n=9$, then $n+d+q=12>9$, $p$ negative, invalid. $d=5$ gives $n=0$, invalid.

- $q=1$: $4n+9d=69$. Again $d\equiv1\pmod4$. $d=1$ gives $n=15$, invalid. $d=5$ gives $n=6$, then $n+d+q=12>9$, invalid.

Hence the only feasible solution with 9 coins has $d=1$.
Larger $d$ forces $n$ too large and $p$ negative.

ANSWER 1: A

Problem 2:
We place a $1.5\times1.5$ square on a $1\times1$ grid. It can be rotated. How many unit squares can meet it in positive area?

If sides are parallel to the grid, an interval of length 1.5 meets at most 3 unit intervals, e.g. $0.9$ to $2.4$ meets $[0,1],[1,2],[2,3]$. So at most $3\times3=9$. This tempts choice C.

But rotation is allowed. The diagonal is $1.5\sqrt2\approx2.12$.
Place the square at $45^\circ$ centered at a lattice point.
Half-diagonal $=1.5/\sqrt2\approx1.06$.
Region: $|x|+|y|\le1.06$ about the center.

$x$-range about $-1.06$ to $1.06$ meets 4 columns: $[-2,-1],[-1,0],[0,1],[1,2]$. Similarly 4 rows.

The center 4 squares are fully met. To the east, the sliver $1<x\le1.06$ near $y=0$ meets both $[1,2]\times[0,1]$ and $[1,2]\times[-1,0]$. Similarly west 2, north 2, south 2. Total $4+8=12$.

The 4 far corner $2\times2$ blocks e.g. $[1,2]\times[1,2]$ need $|x|+|y|\ge2>1.06$, so are missed. So 12 is attained. More than 12 would require hitting a corner block, which is impossible with this small size; 12 is the max.

Hence at least 12 can be attained, so the maximum is in "12 or more."

ANSWER 2: E

Problem 3:
Cards show P, Q, 3, 4, 6. Each has letter on one side, number on the other.
Jane claims: If vowel on one side, then even on the other. i.e. Vowel $\implies$ Even.

To prove Jane wrong we need a counterexample: a card with Vowel on one side AND Odd on the other.

Which face-up cards could hide such a counterexample?

- Showing even (4,6): the back could be vowel or consonant; vowel+even is consistent with the rule, so cannot disprove.
- Showing consonant (P,Q are consonants, vowels are A,E,I,O,U): the back could be odd or even; consonant with anything is consistent, since rule says nothing about consonants.
- Showing odd (3): if the back is a vowel, we have vowel+odd, disproving the rule. If back is consonant, rule holds.

There is no face-up vowel. So the only card that could disprove is the odd number 3. Mary must have turned 3 and found a vowel.

ANSWER 3: A

Problem 4:
Steph: first half $15/20=75\%$, second half $10/10=100\%$. Total $25/30=5/6$.

Candace: let first be $x/12$, second $y/18$, $x,y$ integers.
Steph higher each half:
$$x/12 < 0.75 \implies x<9 \implies x\le8,$$
$$y/18 <1 \implies y\le17.$$

Same overall percentage:
$$\frac{x+y}{12+18}=\frac{25}{30}$$
$$\frac{x+y}{30}=\frac{25}{30}\implies x+y=25.$$

This is Simpson's paradox. From $x\le8$, $y=25-x\ge17$. But $y\le17$. Hence $y=17$, $x=8$.

Difference second minus first:
$$y-x=17-8=9.$$

ANSWER 4: C

Problem 5:
Two dice, $36$ ordered outcomes. Count product $>10$.

- First die 1: products $1$ to $6$, none $>10$: 0.
- 2: need $>5$ on second: only 6: 1.
- 3: need $>10/3$: 4,5,6: 3.
- 4: need $>2.5$: 3,4,5,6: 4.
- 5: need $>2$: 3,4,5,6: 4.
- 6: need $>1.66$: 2,3,4,5,6: 5.

Total $0+1+3+4+4+5=17$.

Probability $=17/36$. This is just under $1/2=18/36$, ruling out C.

ANSWER 5: B

Problem 6:
All numbers use digits 2,4,5,7 once, sum $=18$, so all are multiples of 9. They lie between 2457 and 7542. If one divides another, the ratio is integer $<7542/2457\approx3.07$, so ratio $2$ or $3$.

All choices are $\ge5724>5000$. Doubling exceeds 9999, so none can be the smaller number; each can only be the larger number. So test $N/2$ and $N/3$ for being a permutation of 2,4,5,7.

A. $5724/2=2862$ (has 8,6), $5724/3=1908$ no.
B. $7245$ odd, $7245/3=2415$ (has 1) no.
C. $7254/2=3627$ (has 3,6), $7254/3=2418$ no.
D. $7425/3=2475$, which is exactly digits 2,4,7,5! Indeed $2475\times3=7425$. This works.
E. $7542/2=3771$ no, $7542/3=2514$ (has 1) no.

Thus the unique multiple is 7425, triple of 2475.

ANSWER 6: D

Problem 7:
Let $a_n=a_1+(n-1)d$, $d>0$ integer, $a_1$ integer.

$1\le a_1\le10$,
$13\le a_1+d\le20$,
$241\le a_1+14d\le250$.

From the last, $14d=a_{15}-a_1$. Since $a_1\le10$, $14d\ge241-10=231$. Since $a_1\ge1$, $14d\le250-1=249$.
$$231\le14d\le249$$
$$16.5\le d\le17.78.$$
So $d=17$.

Then $a_{15}=a_1+238\in[241,250]\implies a_1\in[3,12]$. With $a_1\le10$, $a_1\in[3,10]$.

Also $a_2=a_1+17\in[13,20]\implies a_1\le3$. Hence $a_1=3$.

Check: $3,20,\dots,241$ works.

Then $a_{14}=a_1+13d=3+13\cdot17=3+221=224$.
Digit sum $=2+2+4=8$.

ANSWER 7: A

Problem 8:
Day 1 needs $m$ min/mile, then $m+5,m+10,m+15$, $m$ integer. One hour $=60$ min, so miles $=60/(m+5k)$ must be integer. Hence each of $m,m+5,m+10,m+15$ divides 60.

Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.

Need four in arithmetic progression step 5. Try:
$m=5$: 5,10,15,20 all divide 60. Works.
$m=1$: 1,6 divide but 11 fails.
$m=10$: 10,15,20 divide but 25 fails.
All others fail similarly. So $m=5$ uniquely.

Distances: $60/5=12$, $60/10=6$, $60/15=4$, $60/20=3$.
Total $=12+6+4+3=25$.

ANSWER 8: C

Problem 9:
Start 128.

25% to Jill: $0.25\times128=32$ sold. Left $128-32=96$.

25% of remainder to June: $0.25\times96=24$ sold. Left $96-24=72$.

Give 1 away: $72-1=71$ left.

Common error is forgetting the final $-1$ (72) or taking 25% of original twice.

ANSWER 9: D

Problem 10:
Start with 1. Remove $1/2$, leave $1/2$.
Remove $1/3$ of remainder, leave $2/3$ of it: $(1/2)(2/3)=1/3$.
Remove $1/4$ of remainder, leave $3/4

Problem 1:
We are given that there are 9 coins consisting of pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25). The total value is $1.02 (102 cents), and there is at least one coin of each type. We want to find the number of dimes.

Let $p, n, d, q$ denote the number of pennies, nickels, dimes, and quarters, respectively.
1. Total number of coins:
$$p + n + d + q = 9$$
2. Total value in cents:
$$p + 5n + 10d + 25q = 102$$
3. Each variable is an integer with $p, n, d, q \ge 1$.

Notice that $5n + 10d + 25q$ is a multiple of $5$. Therefore, $p$ must have the same remainder as $102$ when divided by $5$, meaning $p$ must end in $2$ or $7$:
- If $p = 7$, then $n + d + q = 9 - 7 = 2$. But since $n \ge 1, d \ge 1, q \ge 1$, we must have $n + d + q \ge 3$, which is impossible.
- Therefore, $p = 2$.

Substitute $p = 2$ into our equations:
$$n + d + q = 7$$
$$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$

Subtract the first equation from the second:
$$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
$$d + 4q = 13$$

Since $q \ge 1$ and $n \ge 1$:
- If $q = 1$, then $d = 9$, which gives $n = 7 - 9 - 1 = -3$, impossible.
- If $q = 2$, then $d = 13 - 8 = 5$, which gives $n = 7 - 5 - 2 = 0$, impossible since $n \ge 1$.
- If $q = 3$, then $d = 13 - 12 = 1$, which gives $n = 7 - 1 - 3 = 3 \ge 1$. This works!
- If $q \ge 4$, then $4q \ge 16 > 13$, which gives no positive solutions for $d$.

Thus, there must be $1$ dime.

ANSWER 1: A

---

Problem 2:
We are given a checkerboard of $1$-inch squares and a square card measuring $1.5$ inches by $1.5$ inches. We want to find the maximum number of squares $n$ that the card can cover (partially or fully).

If the card is placed with its edges parallel to the grid lines of the checkerboard:
- Along each direction, an interval of length $1.5$ can intersect at most $3$ consecutive unit intervals (for example, $[0.8, 2.3]$ intersects $[0, 1], [1, 2],$ and $[2, 3]$).
- Thus, in an axis-aligned position, the card can cover parts of at most $3 \times 3 = 9$ squares.

However, the card does not have to be parallel to the grid lines. Consider rotating the card by $45^\circ$:
- The side length of the card is $s = 1.5$.
- The distance from the center to each of the four vertices is:
  $$\frac{s}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} \approx 1.0607 \text{ inches}$$
- Place the center of this tilted square at the intersection of grid lines, $(0,0)$. The four vertices are at approximately $(0, 1.0607)$, $(1.0607, 0)$, $(0, -1.0607)$, and $(-1.0607, 0)$.
- The four central squares meeting at $(0,0)$—namely $[-1, 0] \times [0, 1]$, $[0, 1] \times [0, 1]$, $[-1, 0] \times [-1, 0]$, and $[0, 1] \times [-1, 0]$—are all clearly covered.
- At the top vertex $(0, 1.0607)$, the card crosses the line $y = 1$ and extends into both $[0, 1] \times [1, 2]$ and $[-1, 0] \times [1, 2]$, covering an area of positive size in each of these $2$ squares.
- By symmetry, at each of the other three vertices (right, bottom, left), the card similarly extends past the boundary to cover parts of $2$ additional squares.

Counting all these squares:
$$4 \text{ (central)} + 4 \times 2 \text{ (outer)} = 12 \text{ squares}$$

Since $12$ squares can be covered, the maximum possible value of $n$ is $12$ or more.

ANSWER 2: E

---

Problem 3:
Jane stated: "If a vowel is on one side of any card, then an even number is on the other side." We want to find which card Mary turned over to prove Jane wrong.

In logic, a conditional statement "If $P$, then $Q$" is only falsified by a counterexample where $P$ is true and $Q$ is false. Here:
- $P$: "The card has a vowel on one side."
- $Q$: "The card has an even number on the other side."

A counterexample must be a card with a **vowel** on one side and an **odd number** on the other side.
- Cards showing consonants ($P$ and $Q$): The statement says nothing about cards with consonants, so turning them over cannot disprove the claim.
- Cards showing even numbers ($4$ and $6$): Even if the other side has a consonant, the statement is not violated because the rule does not require even numbers to only pair with vowels.
- Card showing an odd number ($3$): If the other side of this card is a vowel, then we have a card with a vowel on one side and an odd number on the other, directly disproving Jane's statement.

Thus, Mary turned over card $3$.

ANSWER 3: A

---

Problem 4:
We are given:
- Steph scored $15$ out of $20$ baskets in the first half ($75\%$).
- Steph scored $10$ out of $10$ baskets in the second half ($100\%$).
- Steph's total: $15 + 10 = 25$ baskets out of $20 + 10 = 30$ attempts, so her overall percentage is $\frac{25}{30} = \frac{5}{6}$.
- Candace took $12$ attempts in the first half and $18$ in the second half, for a total of $30$ attempts.
- Candace ended with the same overall percentage as Steph, so Candace made $\frac{5}{6} \times 30 = 25$ baskets in total.

Let $c_1$ and $c_2$ be the number of baskets Candace scored in the first and second halves, respectively. Then:
$$c_1 + c_2 = 25$$

In each half, Steph scored a higher percentage than Candace:
1. First half:
   $$\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} = \frac{9}{12} \implies c_1 < 9 \implies c_1 \le 8$$
2. Second half:
   $$\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$$

From $c_1 \le 8$, we have:
$$c_2 = 25 - c_1 \ge 25 - 8 = 17$$
Since $c_2 \le 17$, we must have $c_2 = 17$, which uniquely forces $c_1 = 8$.

We are asked how many more baskets Candace scored in the second half than in the first:
$$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 4: C

---

Problem 5:
Two standard six-sided dice are tossed, yielding $6 \times 6 = 36$ equally likely outcomes. We want to find the probability that the product of the numbers is greater than $10$.

It is easier to count the complementary outcomes where the product is less than or equal to $10$:
- If the first die is $1$: the second die can be $1, 2, 3, 4, 5, 6$ (all products $\le 6 \le 10$) $\implies 6$ outcomes.
- If the first die i

## Problem 1

We need pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), 9 coins total, worth 102¢, with at least one of each.

Let p, n, d, q be the counts:
- p + n + d + q = 9
- p + 5n + 10d + 25q = 102 (in cents)

Subtracting the first equation from the second (in cents): 4n + 9d + 24q = 93.

**Try q = 3:** then 4n + 9d = 93 − 72 = 21. Since 9d ≡ 21 (mod 4), i.e. d ≡ 1 (mod 4), try d = 1: 4n = 12, so n = 3. Then p = 9 − 1 − 3 − 3 = 2.

**Check:** 2 pennies + 3 nickels + 1 dime + 3 quarters = 2 + 15 + 10 + 75 = 102¢ ✓, and 2+3+1+3 = 9 coins ✓, with at least one of each ✓.

**Check other values of q to confirm uniqueness:** q = 1 gives 4n + 9d = 69 (no valid solution with p ≥ 1); q = 2 gives 4n + 9d = 45 (leads to p ≤ 0); q = 4 gives 4n + 9d = −3 (impossible); q = 0 gives 4n + 9d = 93 (no valid solution). So the only solution is 2 pennies, 3 nickels, 1 dime, 3 quarters.

ANSWER 1: A

## Problem 2

Place the 1.5-inch square **tilted 45°** (as a "diamond"), centered at the center of a 3×3 block of one-inch squares.

The tilted square's horizontal and vertical span is 1.5·√2 ≈ 2.12 inches, so it extends past 2 grid lines in each direction — touching parts of squares in **3 columns and 3 rows**. Its boundary is |x| + |y| ≤ 1.06 (half-diagonal ≈ 1.06). The point (1,1), a corner of the 3×3 block, satisfies |1| + |1| = 2... but measuring from the block's corner squares: the point 1 inch over and 1 inch up from a block corner has |x|+|y| = 1 ≤ 1.06, so the diamond reaches into **all 9** squares (middle and edge squares are reached easily; corner squares are reached since the diamond's boundary passes through the point at distance (0.5, 0.5) from center diagonally).

So n = 9 is achievable.

**Can we beat 9?** The square's diameter is 1.5√2 ≈ 2.12 < 3, so its span in any direction is less than 3 inches — it can cross at most 2 grid lines in each direction, hence touch at most 3 columns and 3 rows: at most 3 × 3 = 9 squares. So the maximum is 9, which falls in "8 or 9."

ANSWER 2: C

## Problem 3

Jane's claim: "If a vowel is on one side of a card, then an even number is on the other side." A counterexample requires a card with a **vowel on one side and an odd number on the other**.

Check each card:
- **4 and 6:** Even numbers are showing. Whatever is on the other side, the claim "vowel → even" is satisfied — these can never disprove Jane.
- **P:** P is a consonant. No vowel is involved, so no violation is possible.
- **3:** The other side is a letter; a violation would need a vowel there — possible, but the visible side gives no vowel to test.
- **Q:** Q is the only **visible vowel**. Jane's claim directly asserts the other side of this card is even. Mary turns the Q over, finds an **odd number**, and Jane's statement is shown false with this single card.

The direct test of Jane's claim is the vowel card, Q.

ANSWER 3: E

## Problem 4

**Steph:** first half 15/20 = 75%, second half 10/10 = 100%. Overall: 25 baskets in 30 attempts = 5/6 ≈ 83.3%.

**Candace:** 12 attempts first half, 18 second half, 30 total. Same overall percentage means she also scored **25 baskets**.

Let x = first-half baskets, y = second-half baskets, x + y = 25.

Conditions (Steph's percentage was higher in each half):
- First half: x/12 < 15/20 = 3/4 → x < 9 → **x ≤ 8**
- Second half: y/18 < 1 → **y ≤ 17** → x = 25 − y ≥ 8

So x = 8 and y = 17.

**Check:** 8/12 ≈ 66.7% < 75% ✓; 17/18 ≈ 94.4% < 100% ✓; overall (8+17)/30 = 25/30 = Steph's 25/30 ✓.

Difference: 17 − 8 = **9**.

ANSWER 4: C

## Problem 5

There are 36 equally likely outcomes. Count pairs (a, b) with ab > 10:

- a = 2: need b > 5 → b = 6: **1** outcome
- a = 3: need b ≥ 4 → b = 4, 5, 6: **3** outcomes
- a = 4: need b ≥ 3 (4·3 = 12 > 10): **4** outcomes
- a = 5: need b ≥ 3 (5·2 = 10 is NOT greater than 10): **4** outcomes
- a = 6: need b ≥ 2 (6·2 = 12): **5** outcomes

Total: 1 + 3 + 4 + 4 + 5 = 17 favorable outcomes.

Probability = 17/36. (Tempting wrong answer 1/2 = 18/36 forgets that 2·5 = 10 doesn't count, or miscounts.)

ANSWER 5: B

## Problem 6

All 24 numbers have digit sum 2+4+5+7 = 18, so every one is divisible by 9. If one of these numbers is a multiple of another, say N = k·M, then since both ≡ 0 (mod 9), and the numbers are 4-digit, the possible multipliers are small: k = 2 or k = 3 (k ≥ 4 would make N ≥ 8000, impossible with digits ≤ 7).

Test each choice:
- **A. 5724:** 5724 ÷ 2 = 2862 — wrong digits; 5724 ÷ 3 = 1908 — wrong digits. ✗
- **B. 7245:** 7245 ÷ 2 not whole; ÷ 3 = 2415 — contains a 1, wrong digits. ✗
- **C. 7254:** ÷ 2 = 3627 ✗; ÷ 3 = 2418 ✗. ✗
- **D. 7425:** 7425 ÷ 3 = **2475** — digits 2, 4, 7, 5, exactly our set! ✓ (2475 is one of the 24 numbers.)
- **E. 7542:** ÷ 2 = 3771 ✗; ÷ 3 = 2514 — contains a 1, ✗. ✗

So 7425 = 3 × 2475.

ANSWER 6: D

## Problem 7

The integers are equally spaced with common difference d, so a₂ = a₁ + d and a₁₅ = a₁ + 14d.

**Bound d using a₁₅:** 241 ≤ a₁ + 14d ≤ 250. Since 1 ≤ a₁ ≤ 10:
- 14d ≥ 241 − 10 = 231 → d ≥ 16.5 → **d ≥ 17**
- 14d ≤ 250 − 1 = 249 → d ≤ 17.78 → **d ≤ 17**

So **d = 17**.

**Find a₁:** a₁ + 14(17) = a₁ + 238 ∈ [241, 250] → a₁ ∈ [3, 12]. Also a₂ = a₁ + 17 ∈ [13, 20] → a₁ ≤ 3. Combined with a₁ ≥ 1: **a₁ = 3**.

Then a₁₅ = 3 + 238 = 241 ✓ (in range), and

a₁₄ = a₁₅ − d = 241 − 17 = **224**.

Sum of digits: 2 + 2 + 4 = **8**.

ANSWER 7: A

## Problem 8

Linda travels 1 mile in m minutes, so in 60 minutes she travels 60/m miles. Her four days use m, m+5, m+10, m+15 minutes per mile, and each distance 60/m, 60/(m+5), 60/(m+10), 60/(m+15) must be a whole number of miles.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. We need four divisors spaced 5 apart, all at most 60.

**Test m:** m = 1 → (1,6,11,16): 11, 16 fail. m = 2 → (2,7,12,17): fail. m = 3 → (3,8,13,18): fail. m = 4 → (4,9,14,19): fail. **m = 5 → (5, 10, 15, 20): all divide 60! ✓** (m = 10 → (10,15,20,25): 25 fails; all larger m fail similarly.)

Distances: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = **25 miles**

Problem 1:  
Let the numbers of pennies, nickels, dimes, and quarters be \(p,n,d,q\), respectively. Then

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Modulo \(3\), this gives \(n\equiv 0\pmod 3\). Since \(n\) is positive and at most \(6\), we have \(n=3\) or \(n=6\).

- If \(n=6\), then

\[
24+9d+24q=93 \implies 3d+8q=23.
\]

The positive integer solution is \(q=1,\ d=5\), but then

\[
p=9-6-5-1=-3,
\]

impossible.

- If \(n=3\), then

\[
12+9d+24q=93 \implies 3d+8q=27.
\]

This gives \(q=3,\ d=1\), and then \(p=2\), which works.

Thus the number of dimes is \(1\).

ANSWER 1: A

Problem 2:  
A square of side length \(1.5\) has diagonal length

\[
1.5\sqrt2=\frac{3\sqrt2}{2}\approx 2.121.
\]

So in any horizontal or vertical direction, its projection has length at most \(2.121<3\). Therefore it can intersect at most \(3\) columns and at most \(3\) rows of unit squares, so it can cover at most

\[
3\cdot 3=9
\]

squares.

This bound is attainable: rotate the square \(45^\circ\). For example, place its center at \((1.5,1.5)\). Its vertices are at distance \(\frac{3\sqrt2}{4}\approx 1.0607\) from its center, and its region is

\[
|x-1.5|+|y-1.5|\le \frac{3\sqrt2}{4}.
\]

Each of the nine squares in the \(3\times3\) block has a point whose distance from the center is at most \(1\), which is strictly less than \(1.0607\). Hence each of those nine squares has positive area covered.

Thus the maximum possible value of \(n\) is \(9\), which is in the choice range “8 or 9.”

ANSWER 2: C

Problem 3:  
Jane’s statement is:

> If one side of a card is a vowel, then the other side is an even number.

A counterexample must have a vowel on one side and an odd number on the other side.

The visible cards are:

\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- \(P\) and \(Q\) are consonants.
- \(4\) and \(6\) are even numbers, so even if their hidden sides are vowels, they do not contradict Jane.
- \(3\) is odd. If the hidden side of \(3\) is a vowel, then we have a vowel on one side and an odd number on the other side, disproving Jane.

Therefore Mary must have turned over the card showing \(3\).

ANSWER 3: A

Problem 4:  
Steph’s totals:

\[
15+10=25 \text{ baskets}, \qquad 20+10=30 \text{ attempts}.
\]

So Steph’s overall percentage is

\[
\frac{25}{30}.
\]

Let Candace have scored \(a\) baskets out of \(12\) attempts in the first half, and \(b\) baskets out of \(18\) attempts in the second half.

Because Steph’s percentage was higher in each half:

\[
\frac{a}{12}<\frac{15}{20}=\frac34 \implies a<9 \implies a\le 8,
\]

and

\[
\frac{b}{18}<\frac{10}{10}=1 \implies b<18 \implies b\le 17.
\]

Their overall percentages are equal, so

\[
\frac{a+b}{30}=\frac{25}{30}\implies a+b=25.
\]

With \(a\le8\) and \(b\le17\), the only possibility is

\[
a=8,\qquad b=17.
\]

Thus Candace scored

\[
17-8=9
\]

more baskets in the second half than in the first half.

ANSWER 4: C

Problem 5:  
There are \(36\) equally likely ordered outcomes for two dice.

Count the outcomes where the product is at most \(10\):

- If the first die is \(1\): \(6\) outcomes.
- If the first die is \(2\): products \(2,4,6,8,10\) give \(5\) outcomes.
- If the first die is \(3\): \(3,6,9\) give \(3\) outcomes.
- If the first die is \(4\): \(4,8\) give \(2\) outcomes.
- If the first die is \(5\): \(5,10\) give \(2\) outcomes.
- If the first die is \(6\): \(6\) gives \(1\) outcome.

Total with product at most \(10\):

\[
6+5+3+2+2+1=19.
\]

So the number with product greater than \(10\) is

\[
36-19=17.
\]

Thus the probability is

\[
\frac{17}{36}.
\]

ANSWER 5: B

Problem 6:  
Every number is a permutation of \(2,4,5,7\), so the smallest is \(2457\) and the largest is \(7542\).

If one of these numbers is a multiple of another distinct number, their ratio must be an integer less than

\[
\frac{7542}{2457}<4.
\]

So the ratio is either \(2\) or \(3\).

Check ratio \(3\): the smaller number must satisfy

\[
\text{smaller}\le \frac{7542}{3}=2514.
\]

The only permutations in that range are \(2457\) and \(2475\).

\[
3\cdot2457=7371
\]

is not a permutation of \(2,4,5,7\), but

\[
3\cdot2475=7425
\]

is exactly one of the allowed numbers.

Check ratio \(2\): the smaller number must be at most \(3771\), so it must begin with \(2\). The possibilities are

\[
2457,2475,2547,2574,2745,2754.
\]

Their doubles are

\[
4914,4950,5094,5148,5490,5508,
\]

none of which are permutations of \(2,4,5,7\).

Thus the only number that is a multiple of another is

\[
7425.
\]

ANSWER 6: D

Problem 7:  
Let the common difference be \(d\), and let \(a_1=a\). Then

\[
a_2=a+d,\qquad a_{15}=a+14d.
\]

We are given

\[
1\le a\le10,
\]
\[
13\le a+d\le20,
\]
\[
241\le a+14d\le250.
\]

From the last condition and \(1\le a\le10\),

\[
241-10\le14d\le250-1,
\]

so

\[
231\le14d\le249.
\]

Thus \(14d=238\), giving

\[
d=17.
\]

Now

\[
13\le a+17\le20 \implies -4\le a\le3.
\]

Together with \(1\le a\le10\), this gives \(1\le a\le3\).

Also,

\[
241\le a+238\le250 \implies 3\le a\le12.
\]

Together, this forces

\[
a=3.
\]

Then

\[
a_{14}=a+13d=3+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:  
Let \(m\) be the number of minutes per mile on the first day. Since Linda traveled for one hour each day, the distances are

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\quad \frac{60}{m+15}.
\]

Each of these must be an integer number of miles, so each denominator must divide \(60\).

The positive divisors of \(60\) are

\[
1,2,3,4,5,6,10,12,15,20,30,60.
\]

We need four divisors spaced by \(5\). The only such set is

\[
5,10,15,20.
\]

Then the distances are

\[
\frac{60}{5}=12,\quad \frac{60}{10}=6,\quad \frac{60}{15}=4,\quad \frac{60}{20}=3.
\]

Their sum is

\[
12+6+4+3=25.
\]

ANSWER 8: C

Problem 9:  
Jack starts with \(128\) apples.

After selling \(25\%\) to Jill, he has

\[
128\left(\frac34\right)=96
\]

apples.

After selling \(25\%\) of the remainder to June, he has

Problem 1:

We need determine the number of dimes among 9 coins totaling \(102\) cents.

Let

- \(p=\) number of pennies,
- \(n=\) number of nickels,
- \(d=\) number of dimes,
- \(q=\) number of quarters.

We know

\[
p+n+d+q=9
\]

and

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there are only 9 coins and at least one of each type, \(q\) cannot be very large.

Try possible values of \(q\):

- If \(q=1\), then
  \[
  4n+9d=69.
  \]
  This gives \(d=5,n=6\) as the smallest positive possibility, already totaling more than 9 coins.

- If \(q=2\), then
  \[
  4n+9d=45.
  \]
  The positive possibility \(d=1,n=9\) again uses too many coins.

- If \(q=3\), then
  \[
  4n+9d=21.
  \]
  Taking \(d=1\) gives
  \[
  4n=12,\qquad n=3.
  \]
  Then
  \[
  p=9-(3+3+1)=2.
  \]

Indeed, \(2\) pennies, \(3\) nickels, \(1\) dime, and \(3\) quarters total

\[
2+15+10+75=102.
\]

So there must be exactly \(1\) dime. Choices B through E would not allow a valid collection of only 9 coins.

ANSWER 1: A

Problem 2:

We want the greatest possible number of one-inch checkerboard squares that a \(1.5\)-inch square card can cover at least partly.

Rotate the card \(45^\circ\), making it look like a diamond. The horizontal and vertical width of the rotated card is

\[
1.5\sqrt2 \approx 2.12\text{ inches}.
\]

Place the center of this diamond at a checkerboard grid intersection.

The diamond covers:

- the 4 unit squares meeting at the center grid intersection, and
- 2 more squares in each of the four directions: left, right, up, and down.

Thus it covers at least

\[
4+2+2+2+2=12
\]

checkerboard squares.

Therefore the maximum is at least 12, so choices A through D, all of which say at most 11, cannot be correct.

ANSWER 2: E

Problem 3:

Jane’s statement is:

> If a vowel is on one side of a card, then an even number is on the other side.

To prove Jane wrong, Mary must find a card having:

- a vowel on one side, and
- an odd number on the other side.

The visible odd-numbered card is \(3\). If the other side of the \(3\) card is a vowel, then Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the statement, because they are even numbers, which are allowed to have vowels on the other side.

Turning over \(P\) or \(Q\) cannot help because those are consonants, not vowels.

Therefore Mary turned over the card showing \(3\).

ANSWER 3: A

Problem 4:

Steph made

\[
15+10=25
\]

baskets in

\[
20+10=30
\]

attempts. Her overall percentage was

\[
\frac{25}{30}=\frac56.
\]

Candace also had \(12+18=30\) attempts, and her overall percentage was the same as Steph’s. Therefore Candace also made

\[
\frac56\cdot 30=25
\]

baskets total.

Let Candace make \(x\) baskets in the first half. Then she made \(25-x\) baskets in the second half.

In the first half, Steph’s percentage was

\[
\frac{15}{20}=75\%.
\]

Candace’s percentage was lower, so

\[
\frac{x}{12}<\frac{15}{20}.
\]

Thus

\[
x<9.
\]

In the second half, Steph made all her shots:

\[
\frac{10}{10}=100\%.
\]

Candace had a lower percentage, so she could not make all 18 shots:

\[
25-x<18.
\]

Thus

\[
x>7.
\]

The only whole number satisfying

\[
7<x<9
\]

is

\[
x=8.
\]

So Candace made

\[
25-8=17
\]

baskets in the second half. The difference is

\[
17-8=9.
\]

ANSWER 4: C

Problem 5:

There are

\[
6\cdot 6=36
\]

equally likely ordered outcomes when two dice are tossed.

Count the outcomes for which the product is greater than 10.

- First die \(=1\): no products exceed 10.
- First die \(=2\): only \(2\cdot 6=12\), so \(1\) outcome.
- First die \(=3\): second die can be \(4,5,6\), so \(3\) outcomes.
- First die \(=4\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=5\): second die can be \(3,4,5,6\), so \(4\) outcomes.
- First die \(=6\): second die can be \(2,3,4,5,6\), so \(5\) outcomes.

Total favorable outcomes:

\[
1+3+4+4+5=17.
\]

Therefore the probability is

\[
\frac{17}{36}.
\]

For example, \((2,5)\) is not counted because its product is exactly \(10\), not greater than \(10\).

ANSWER 5: B

Problem 6:

Each number uses digits \(2,4,5,7\), whose sum is

\[
2+4+5+7=18.
\]

The choices are all fairly large numbers. Since the smallest possible number is \(2457\), a chosen number larger than \(5000\) cannot be doubled and still remain a four-digit rearrangement of these digits. So check whether each choice is divisible by \(2\) or \(3\) to produce another valid rearrangement.

- \(5724/2=2862\), not a rearrangement of \(2,4,5,7\).
- \(7245/3=2415\), not a rearrangement.
- \(7254/2=3627\), not a rearrangement.
- \(7425/3=2475\), which does use exactly the digits \(2,4,5,7\).
- \(7542/3=2514\), not a rearrangement.

Indeed,

\[
2475\cdot 3=7425.
\]

Thus \(7425\) is the one that is a multiple of another number in the list of permutations.

ANSWER 6: D

Problem 7:

The integers are equally spaced, so they form an arithmetic sequence. Let the common difference be \(d\).

We are given

\[
13\le a_2\le20
\]

and

\[
241\le a_{15}\le250.
\]

Since \(a_{15}\) is 13 steps after \(a_2\),

\[
a_{15}=a_2+13d.
\]

Therefore

\[
241-20\le 13d\le250-13.
\]

So

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

Thus \(d\) is either \(17\) or \(18\).

If \(d=18\), then

\[
a_{15}=a_1+14(18)=a_1+252,
\]

which is greater than \(250\), since \(a_1\ge1\). So \(d\neq18\).

Thus

\[
d=17.
\]

Now

\[
a_{15}=a_1+14(17)=a_1+238.
\]

Since \(a_{15}\ge241\),

\[
a_1+238\ge241,
\]

so \(a_1\ge3\).

Also,

\[
a_2=a_1+17\le20,
\]

so \(a_1\le3\). Therefore

\[
a_1=3.
\]

Then

\[
a_{14}=a_1+13(17)=3+221=224.
\]

The sum of its digits is

\[
2+2+4=8.
\]

ANSWER 7: A

Problem 8:

Let \(m\) be the number of minutes Linda takes to travel one mile on the first day.

Her times per mile for the four days are

\[
m,\quad m+5,\quad m+10,\quad m+15.
\]

Since she travels for one hour, or 60 minutes, each day’s distance is

\[
\frac{60}{m},\quad \frac{60}{m+5},\quad \frac{60}{m+10},\qu