🔬 Model Lab

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,

Problem 1:
We need four-digit numbers using 2,4,5,7 once each.
There are 24 of them. Sum of digits = 2+4+5+7 = 18,
so every such number is divisible by 9.

Let a smaller one $S$ divide a larger one $L$, $L=kS$.
Since $2457 \le S < L \le 7542$,
$$ k = L/S < 7542/2457 < 3.07. $$
$k$ is an integer $>1$, so $k=2$ or $k=3$.

Rule out $k=2$.
$S\ge 2457$ and $L=2S\le 7542$ gives $S\le 3771$.
Hence $S$ is $2457$ to $3771$ and uses digits 2,4,5,7,
so its thousands digit must be 2. So $S=2\_\_\_$.

Let $d$ be units of $S$. $d$ is 4,5 or 7.
$L=2S$ has units $(2d)\bmod 10$, which must be 2,4,5 or 7.
- $d=4 \to 8$, impossible.
- $d=5 \to 0$, impossible.
- $d=7 \to 4$, possible, with carry 1.

So $S$ must end in 7, $L$ end in 4, and $S=2\_\_7$
with middle digits 4,5. I.e. $S=2457$ or $2547$.
But $2457\times 2=4914$, $2547\times 2=5094$,
neither uses digits 2,4,5,7. So $k=2$ is impossible.

Hence $k=3$. Then $L=3S\le 7542$ gives $S\le 2514$.
Also $S\ge 2457$. So $2457\le S\le 2514$ using digits 2,4,5,7.
Thousands digit 2. The only permutations in that interval are
2457 and 2475; $2547$ is already too big since $3\times2547>7542$.

Test:
$2457\times 3 = 7371$, not a permutation.
$2475\times 3 = 7425$, which is $2,4,7,5$ exactly once.

Indeed $2475\times 3 = 6000+1200+210+15 = 7425$.
So the unique multiple is 7425. This eliminates A,B,C,E; e.g.
halves and thirds of 5724,7245,7254,7542 are not permutations.
ANSWER 1: D

Problem 2:
We multiply $N=99\dots9$ ($n$ nines) by $M=44\dots4$ ($n$ fours).
Test small $n$:
$9\times4=36$, digit sum 9.
$99\times44=4356$, digit sum 18.
$999\times444=443556$, digit sum 27.
Guess sum $=9n$.

Prove: $M$ has $n$ digits 4. $N=10^{n}-1$.
$$ N\cdot M = M\cdot10^{n}-M = (M-1)10^{n}+ (10^{n}-M). $$
Let $F=M=44\dots4$. Then
$$M-1 = 44\dots43,$$
$n-1$ fours then 3, and
$$10^{n}-M = 55\dots56,$$
$n-1$ fives then 6, e.g. $1000-444=556$.
Hence the product is the concatenation:
$(n-1$ fours$)3\,(n-1$ fives$)6$.

Digit sum $=4(n-1)+3+5(n-1)+6 =9(n-1)+9=9n$.

For $n=94$, sum $=9\times94=846$.

This rules out 855,945,954,1072.
ANSWER 2: A

Problem 3:
Points per race: 5 first, 3 second, 1 third, 0 otherwise.
Let our student have total $T$. Another student maximized
takes the best place left by ours in each race:
if ours is 1st, max 3; if 2nd,3rd or out, max 5.

$T=13$ can only be $5+5+3$: two wins and a second.
Max for opponent $=3+3+5=11<13$. So 13 guarantees sole first.
Hence 15 also does, but we want smallest.

$T=11$ can be $5+5+1$. Then an opponent can be second, second,
first: $3+3+5=11$, tying us. So 11 does not guarantee strictly
more. It can also be $5+3+3$, where opponent can get
$3+5+5=13>11$.

Thus 11 fails. Any $T\le10$ also fails: e.g. opponent can still
aim for $3+3+5=11$ while we win twice and do worse in the third
race; e.g. we get $5+5+0=10$ and opponent gets 11 and beats us.
Similarly 9 is beatable.

So among attainable totals, 13 is the first that works.
Choices 9,10,11 fail; 15 works but is not smallest.
ANSWER 3: D

Problem 4:
Let $c$ correct, $u$ unanswered, $w$ wrong, $c+u+w=20$.
Score $S=5c+u$.

Then $u=S-5c\ge0$ and $c+u\le20$, i.e. $c+S-5c\le20$,
$S-4c\le20$.

- 90: $u=90-5c$. Need $5c\le90$, $90-4c\le20$.
$c=18$ gives $u=0$, $c+u=18\le20$ with $w=2$. Possible.

- 91: $u=91-5c$. $c=18$ gives $u=1$, $c+u=19\le20$ with $w=1$.
Possible: 18 correct, 1 blank, 1 wrong.

- 92: $u=92-5c$. $c=18$ gives $u=2$, $c+u=20$. Possible:
18 correct, 2 blank.

- 95: $u=95-5c$. $c=19$ gives $u=0$, $c+u=19\le20$ with $w=1$.
Possible: 19 correct, 1 wrong.

- 97: $u=97-5c$. Need $5c\le97$ so $c\le19$, and
$97-4c\le20$ so $4c\ge77$ so $c\ge20$. Impossible.
Indeed $c=19$ would need $u=2$, total $21>20$; $c=20$ gives 100.
So 97 cannot occur.

Hence NOT possible is 97.
ANSWER 4: E

Problem 5:
Cost: 4 for \$5, so per disk $5/4=1.25$.
Sell: 3 for \$5, so per disk $5/3\approx1.67$.

Profit per disk $=5/3-5/4=(20-15)/12=5/12$ dollars.

To make \$100 need $N$ with $N\cdot5/12=100$,
$$N=100\times12/5=240.$$

Check: buy 240 costs $240/4\times5=300$; sell 240 gets
$240/3\times5=400$; profit $100$.
100,120,200,1200 give profits $125/3,50,250/3,500$, not 100.
ANSWER 5: D

Problem 6:
Row 1 has 10 seats, row $i$ has $10+i-1=9+i$ seats.
Row 20 has 29 seats.

In a row of $n$ seats in a line with no two occupied adjacent,
the maximum is to take seats 1,3,5,... i.e. $\lceil n/2\rceil$.

So row $n$ contributes $\lceil n/2\rceil$:
10→5, 11→6, 12→6, 13→7, 14→7, 15→8, 16→8, 17→9,
18→9, 19→10, 20→10, 21→11, 22→11, 23→12, 24→12,
25→13, 26→13, 27→14, 28→14, 29→15.

Summing in pairs: $11+13+15+17+19+21+23+25+27+29$.
This is 10 terms averaging $(11+29)/2=20$, total $200$.

Leaving a seat empty can only reduce total, so 200 is max.
This rules out 150,180,400,460.
ANSWER 6: C

Problem 7:
Let $a_1=A$, common difference $d>0$. Since $a_1,a_2$ integers,
$d=a_2-a_1$ is an integer. $a_n=A+(n-1)d$.

$a_1\in[1,10]$, $a_1+d\in[13,20]$, $A+14d\in[241,250]$.

Then $14d=(A+14d)-A\in[241-10,250-1]=[231,249]$.
So $231\le14d\le249$, $16.5\le d\le17.78$.
Integer $d=17$ only, since $14\times16=224$, $14\times18=252$.

Then $A=(A+14d)-238\in[241-238,250-238]=[3,12]$.
Intersect $[1,10]$ gives $[3,10]$.
Also $A+17\in[13,20]$ gives $A\le3$. Hence $A=3$.

Check: $3,20,\dots,3+14\times17=241$, satisfies all.
Then $a_{14}=A+13d=3+13\times17=3+221=224$.
Digit sum $=2+2+4=8$, ruling out 9,10,11,12.
ANSWER 7: A

Problem 8:
Sum $1+\cdots+9=45$. Three groups equal sum $\Rightarrow$
each sum $45/3=15$.

Find triples of distinct 1–9 summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

9 must be with $\{1,5\}$ or $\{2,4\}$.

Case 9 with 1,5: use $\{1,5,9\}$. Remaining $\{2,3,4,6,7,8\}$.
8 needs two summing to 7: only $3+4$ remains, so $\{3,4,8\}$
forced. Left $\{2,6,7\}=15$. Partition 1.

Case 9 with 2,4: use $\{2,4,9\}$. Remaining $\{1,3,5,6,7,8\}$.
8 needs 7: only $1+6$ remains, so $\{1,6,8\}$ forced.
Left $\{3,5,7\}=15$. Partition 2.

No other cases. So exactly 2 unordered partitions.
Labeled groups would give 12, not liste

### Problem 1:

**Restatement of the Problem:**  
We are given the four digits $2, 4, 5,$ and $7$. There are $4! = 24$ four-digit numbers formed by using each of these digits exactly once. We are told that exactly one of these $24$ numbers is a multiple of another one among them. We need to identify that number from the given choices.

**Step-by-Step Solution:**
1. The smallest possible four-digit number that can be formed using digits $\{2, 4, 5, 7\}$ is $2457$, and the largest is $7542$.
2. If a number $B$ is a multiple of another number $A$, then $B = k \cdot A$ for some integer $k \ge 2$.
3. Since $A \ge 2457$ and $B \le 7542$, the multiplier $k$ can only be:
   $$\frac{7542}{2457} \approx 3.07 \implies k = 2 \text{ or } 3$$
4. Let's test multiples of the smallest valid numbers:
   - For $A = 2457$:
     - $2457 \times 2 = 4914$ (contains the digit $9$, not valid)
     - $2457 \times 3 = 7371$ (contains $3$ and $1$, not valid)
   - For $A = 2475$:
     - $2475 \times 2 = 4950$ (not valid)
     - $2475 \times 3 = 7425$ 
5. Notice that $7425$ contains the digits $7, 4, 2, 5$, each exactly once!
6. Thus, $7425$ is a multiple of $2475$ ($7425 = 3 \times 2475$). 
7. Checking the choices, $7425$ is Choice D.

ANSWER 1: D

---

### Problem 2:

**Restatement of the Problem:**  
Find the sum of the digits of the product of a string of $94$ nines and a string of $94$ fours:
$$\underbrace{999\dots99}_{94\text{ nines}} \times \underbrace{444\dots44}_{94\text{ fours}}$$

**Step-by-Step Solution:**
1. Let $N = \underbrace{444\dots44}_{94\text{ fours}}$.
2. A string of $94$ nines can be written as $10^{94} - 1$.
3. The product is:
   $$(10^{94} - 1) \times N = N \times 10^{94} - N$$
4. Let us observe the pattern for smaller values of $n$ (the number of digits):
   - For $n = 1$: $9 \times 4 = 36$, digit sum is $3 + 6 = 9 = 9 \times 1$.
   - For $n = 2$: $99 \times 44 = 4356$, digit sum is $4 + 3 + 5 + 6 = 18 = 9 \times 2$.
   - For $n = 3$: $999 \times 444 = 443556$, digit sum is $4 + 4 + 3 + 5 + 5 + 6 = 27 = 9 \times 3$.
5. In general, for $n$ digits, the product is:
   $$\underbrace{44\dots4}_{n-1\text{ fours}} 3 \underbrace{55\dots5}_{n-1\text{ fives}} 6$$
   The sum of the digits is:
   $$4(n-1) + 3 + 5(n-1) + 6 = 9(n-1) + 9 = 9n$$
6. Here, $n = 94$, so the sum of the digits is:
   $$9 \times 94 = 846$$

ANSWER 2: A

---

### Problem 3:

**Restatement of the Problem:**  
In each of three races, the first-place finisher receives $5$ points, the second-place finisher receives $3$ points, and the third-place finisher receives $1$ point. There are no ties. We wish to find the smallest number of points a student must earn in the three races to be guaranteed to have strictly more points than any other student.

**Step-by-Step Solution:**
1. In each race, the points awarded are $5, 3,$ and $1$. The possible point values a single student can obtain across $3$ races come from sums of three values chosen from $\{5, 3, 1, 0\}$ (where $0$ represents placing fourth or lower).
2. The possible scores near the top are:
   - $5 + 5 + 5 = 15$
   - $5 + 5 + 3 = 13$
   - $5 + 5 + 1 = 11$
   - $5 + 3 + 3 = 11$
   - $5 + 3 + 1 = 9$ or $3 + 3 + 3 = 9$
3. Can a student earn $11$ points and tie with another student?
   - Yes: Student A could finish 1st, 1st, and 3rd, scoring $5 + 5 + 1 = 11$.
   - Simultaneously, Student B could finish 2nd, 2nd, and 1st, scoring $3 + 3 + 5 = 11$.
   - Since Student A and Student B both have $11$ points, having $11$ points does not guarantee strictly more points than every other student.
4. Next possible higher score is $13$:
   - To score $13$ points, a student must finish with two 1st-place finishes and one 2nd-place finish ($5 + 5 + 3 = 13$).
   - If Student A scores $13$ points, the remaining points available to all other competitors across the three races are:
     - One 1st-place finish ($5$ points),
     - Two 2nd-place finishes ($3$ points each),
     - Three 3rd-place finishes ($1$ point each).
   - The maximum points any other single competitor could obtain is by taking the remaining 1st-place finish and both remaining 2nd-place finishes: $5 + 3 + 3 = 11$ points.
   - Since $11 < 13$, no other student can reach $13$ points.
5. Therefore, $13$ is the smallest number of points that guarantees a student has strictly more points than any other student.

ANSWER 3: D

---

### Problem 4:

**Restatement of the Problem:**  
A test consists of $20$ questions. Correct answers earn $5$ points, unanswered questions earn $1$ point, and incorrect answers earn $0$ points. We need to determine which of the given scores is NOT possible: $90, 91, 92, 95,$ or $97$.

**Step-by-Step Solution:**
1. Let $c$ be the number of correct answers, $u$ be the number of unanswered questions, and $w$ be the number of incorrect answers.
2. We have the constraints:
   $$c + u + w = 20, \quad c, u, w \ge 0$$
   $$\text{Score} = 5c + u$$
3. Let's analyze scores close to $100$:
   - If $c = 20$, then $u = 0, w = 0 \implies \text{Score} = 100$.
   - If $c = 19$, then $u + w = 1$.
     - If $u = 1, w = 0 \implies \text{Score} = 5(19) + 1 = 96$.
     - If $u = 0, w = 1 \implies \text{Score} = 5(19) + 0 = 95$.
   - If $c \le 18$, the maximum possible score is when all remaining questions are left unanswered ($u = 20 - c$):
     $$\text{Score} \le 5(18) + 2 = 92$$
4. Therefore, any score strictly between $92$ and $95$, or between $96$ and $100$, is impossible:
   - Impossible scores in the nineties are: $93, 94, 97, 98, 99$.
5. Checking the choices:
   - $90$: $c = 18, u = 0, w = 2 \implies 5(18) + 0 = 90$ (Possible)
   - $91$: $c = 18, u = 1, w = 1 \implies 5(18) + 1 = 91$ (Possible)
   - $92$: $c = 18, u = 2, w = 0 \implies 5(18) + 2 = 92$ (Possible)
   - $95$: $c = 19, u = 0, w = 1 \implies 5(19) + 0 = 95$ (Possible)
   - $97$: Impossible, as shown above.

ANSWER 4: E

---

### Problem 5:

**Restatement of the Problem:**  
Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. How many disks mus

# Problem 1:
We have all 24 permutations of digits {2, 4, 5, 7}. The smallest is 2457 and the largest is 7542. If one of these numbers is a multiple of another, say X = k·Y, then k = X/Y ≤ 7542/2457 ≈ 3.07, so **k = 2 or k = 3**.

**Try k = 2:** Then 2Y ≤ 7542 forces Y ≤ 3771, so Y starts with 2 or 4, and Y must be even. Candidates and their doubles:
- 2574 → 5148 ✗, 2754 → 5508 ✗
- 4572 → 9144 ✗, 4752 → 9504 ✗

None of the doubles use only digits {2,4,5,7}. So doubling never works.

**Try k = 3:** Then 3Y ≤ 7542 forces Y ≤ 2514, so Y ∈ {2457, 2475}:
- 3 × 2457 = 7371 ✗
- 3 × 2475 = **7425** ✓ (uses exactly digits 2, 4, 5, 7)

So the only multiple pair is 7425 = 3 × 2475. Checking the other choices directly (e.g., 5724 = 2²·3³·53 and 7245 = 3²·5·7·23 have no 4-digit divisors made of 2,4,5,7) confirms none of them works.

ANSWER 1: D

# Problem 2:
A string of 94 nines equals 10⁹⁴ − 1. So the product is

44…4 (94 fours) × (10⁹⁴ − 1) = 44…4000…0 (94 fours followed by 94 zeros) − 44…4.

Look at small cases to see the pattern:
- 44 × 99 = 4356
- 444 × 999 = 443556
- 4444 × 9999 = 44435556

The pattern is: **(n−1) fours, then a 3, then (n−1) fives, then a 6**. (This is the subtraction: the trailing zeros minus 44…4 gives 55…56, and borrowing 1 turns the leading 44…4 into 44…3.)

For n = 94 the answer is: 93 fours, then 3, then 93 fives, then 6.

Sum of digits = 93(4) + 3 + 93(5) + 6 = 372 + 3 + 465 + 6 = **846**.
(Not 954 — that would come from wrongly counting 94 fours and 94 fives.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races. A student's score is a sum of three odd numbers (5, 3, or 1), so every score is **odd**; possible scores are 3, 5, 7, 9, 11, 13, 15.

**Why 13 guarantees the win:** The only way to score 13 is 5 + 5 + 3 (first twice, second once). Then in the two races that student won, any rival scored at most 3 points; in the remaining race, at most 5. So every rival has at most 3 + 3 + 5 = 11 < 13. ✓

**Why smaller values fail:** With 11 points (e.g., 5 + 5 + 1: first twice, third once), another student could be second in those two races (3 + 3) and first in the third (5), also totaling 11 — a tie, so 11 does not guarantee being strictly ahead. (With 9 or 10, a rival could even score more.) Since all scores are odd, 12 is impossible anyway.

ANSWER 3: D

# Problem 4:
Let c = correct, u = unanswered, w = wrong, with c + u + w = 20. Score = 5c + u.

- 90: 18 correct, 0 unanswered, 2 wrong → 18·5 = 90 ✓
- 91: 18 correct, 1 unanswered, 1 wrong → 90 + 1 ✓
- 92: 18 correct, 2 unanswered, 0 wrong → 90 + 2 ✓
- 95: 19 correct, 0 unanswered, 1 wrong → 95 ✓
- 97: Need 5c + u = 97. Since 97 ≡ 2 (mod 5), u ∈ {2, 7, 12, 17}.
  - u = 2 → c = 19, but c + u = 21 > 20 ✗
  - u = 7 → c = 18, c + u = 25 > 20 ✗
  - u = 12 → c = 17, c + u = 29 > 20 ✗
  - u = 17 → c = 16, c + u = 33 > 20 ✗

So 97 is impossible.

ANSWER 4: E

# Problem 5:
Buying at 4 disks for $5 means cost per disk = 5/4 = $1.25.
Selling at 3 disks for $5 means revenue per disk = 5/3 ≈ $1.67.

Profit per disk = 5/3 − 5/4 = (20 − 15)/12 = **5/12 dollars**.

To make $100 profit: number of disks = 100 ÷ (5/12) = 100 · 12/5 = **240**.

Check: 240 disks cost 240 · 1.25 = $300; selling 240 disks brings 240 · 5/3 = $400; profit = $100. ✓
(The tempting wrong answer 120 comes from forgetting she must also *buy* the disks she sells.)

ANSWER 5: D

# Problem 6:
Row i has 10 + (i − 1) seats, so the rows have 10, 11, 12, …, 29 seats.

In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (alternate seating: seat, gap, seat, gap, …).

Compute row by row:
- Row 10 → 5, row 11 → 6, row 12 → 6, row 13 → 7, row 14 → 7, row 15 → 8, row 16 → 8, row 17 → 9, row 18 → 9, row 19 → 10, row 20 → 10, row 21 → 11, row 22 → 11, row 23 → 12, row 24 → 12, row 25 → 13, row 26 → 13, row 27 → 14, row 28 → 14, row 29 → 15.

Total = 5 + (6+6) + (7+7) + (8+8) + (9+9) + (10+10) + (11+11) + (12+12) + (13+13) + (14+14) + 15 = **200**.

(Shortcut: total seats = 390; half is 195; the 10 odd-sized rows each contribute an extra half-seat: 195 + 10 = 200. Choice 400 wrongly assumes every seat can be filled; choice 180 wrongly floors every row.)

ANSWER 6: C

# Problem 7:
Let d be the common spacing. Then a₂ = a₁ + d and a₁₅ = a₁ + 14d.

From 1 ≤ a₁ ≤ 10 and 241 ≤ a₁₅ ≤ 250:
241 ≤ a₁ + 14d ≤ 250 ⟹ 231 ≤ 14d ≤ 249 ⟹ 16.5 ≤ d ≤ 17.79, so **d = 17** (integers only).

Now a₁₅ = a₁ + 14(17) = a₁ + 238 must satisfy 241 ≤ a₁ + 238 ≤ 250, so a₁ ≥ 3.
Also a₂ = a₁ + 17 must satisfy a₂ ≤ 20, so a₁ ≤ 3.
Therefore a₁ = 3. (Check: a₂ = 20 ∈ [13, 20] ✓, a₁₅ = 241 ∈ [241, 250] ✓.)

Then a₁₄ = a₁ + 13d = 3 + 13(17) = 3 + 221 = **224**.

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

ANSWER 7: A

# Problem 8:
The numbers 1 through 9 sum to 45, so each of the 3 groups must sum to 45/3 = **15**.

All triples from {1,…,9} summing to 15:
1-5-9, 1-6-8, 2-4-9, 2-5-8, 2-6-7, 3-4-8, 3-5-7, 4-5-6 (these are the 8 lines of the 3×3 magic square).

Since the group containing 1 must be either {1,5,9} or {1,6,8}:

**Case {1,5,9}:** Remaining digits {2,3,4,6,7,8} must form two triples summing to 15 — the only option is {2,6,7} and {3,4,8}. ✓ → Partition: {1,5,9}, {2,6,7}, {3,4,8}.

**Case {1,6,8}:** Remaining digits {2,3,4,5,7,9} — the only option is {2,4,9} and {3,5,7}. ✓ → Partition: {1,6,8}, {2,4,9}, {3,5,7}.

No other partitions exist, so there are exactly **2** ways (groups aren't labeled, and order within a group doesn't matter).

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count pairs (die1, die2) with product > 10:

- Roll of 1: products 1–6 → none.
- Roll of 2: products 2,4,6,8,10,12 → (2,6) only → 1 outcome.
- Roll of 3: 3,6,9,12,15,18 → (3,4),(3,5),(3,6) → 3 outcomes.
- Roll of 4: 4,8,12,16,20,24 → (4,3),(4,4),(4,5),(4,6) → 4 outcomes.
- Roll of 5: 5,10,15,20,25,30 → (5,3),(5,4),(5,5),(5,6) → 4 outcomes.
- Roll of 6: 6,12,18,24,30,36 → (6,2),(6,3),(6,4),(6,5),(6,6) → 5 outcomes.

Total favorable = 1 + 3 + 4 + 4 + 5 = **17**

# Problem 1:
We need a permutation of {2,4,5,7} that is a multiple of another permutation of the same digits.

The smallest such number is 2457 and the largest is 7542, so the ratio of the two numbers is less than 7542/2457 ≈ 3.07. So the multiplier can only be 2 or 3.

**Multiplier 3:** Try the small permutations starting with 2 (since 3 × (number) must still be a 4-digit permutation):
- 2457 × 3 = 7371 (no)
- 2475 × 3 = **7425** ✓ — digits 7, 4, 2, 5, a permutation!

So 7425 = 3 × 2475, both using 2, 4, 5, 7 exactly once.

Checking the tempting choices: 5724, 7245, 7254, 7542 are all even or not divisible appropriately — e.g. 7254/2 = 3627 and 7254/3 = 2418 (digits wrong), 5724/2 = 2862, 7542/2 = 3771, 7542/3 = 2514, 7245/3 = 2415 — none give permutations of 2,4,5,7.

ANSWER 1: D

# Problem 2:
Look at the pattern with small strings:
- 9 × 4 = 36, digit sum = 9
- 99 × 44 = 4356, digit sum = 4+3+5+6 = 18
- 999 × 444 = 443556, digit sum = 4+4+3+5+5+6 = 27
- 9999 × 4444 = 44435556, digit sum = 36

The product of *n* nines and *n* fours is 44…4 (n−1 fours) 3 55…5 (n−1 fives) 6, whose digit sum is
4(n−1) + 3 + 5(n−1) + 6 = 9n − 9 + 9 = 9n.

For n = 94: digit sum = 9 × 94 = **846**.

(945 and 954 would come from miscounting as 9(n+1) or similar.)

ANSWER 2: A

# Problem 3:
Each race awards 5 + 3 + 1 = 9 points, so 27 points total over three races.

**Is 11 enough?** A student could get 11 as 5 + 5 + 1. Then in the third race someone else could win (5) and place second in the first two races: 3 + 3 + 5 = 11 — a tie, so 11 is not a guarantee. (Similarly 9 and 10 fail.)

**Is 13 enough?** The only way to score 13 is 5 + 5 + 3. Then the remaining points available are: race 1 → {3, 1}, race 2 → {3, 1}, race 3 → {5, 1}. The best any other student can do is 3 + 3 + 5 = 11 < 13. So 13 guarantees the win.

15 (winning all three) works too, but is not the smallest.

ANSWER 3: D

# Problem 4:
Let c = number correct, u = number unanswered, with c + u ≤ 20. Score = 5c + u.

- 90 = 5(18) + 0 → 18 questions used ✓
- 91 = 5(18) + 1 → 19 used ✓
- 92 = 5(18) + 2 → 20 used ✓
- 95 = 5(19) + 0 → 19 used ✓
- 97: need 5c + u = 97. If c = 19, u = 2, but 19 + 2 = 21 > 20 ✗. If c = 18, u = 7, total 25 > 20 ✗. Smaller c makes u even bigger. With c = 20, score is 100. So 97 is impossible.

ANSWER 4: E

# Problem 5:
Cost per disk = $5/4 = $1.25. Selling price per disk = $5/3 ≈ $1.6667.

Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.

Number needed: 100 ÷ (5/12) = 100 × 12/5 = **240** disks.

(Choosing 120 comes from using a profit of $5/6 per disk; 1200 from forgetting the 5.)

ANSWER 5: D

# Problem 6:
Rows have 10, 11, 12, …, 29 seats. In a row of n seats with no two students adjacent, the maximum is ⌈n/2⌉ (seat in positions 1, 3, 5, …).

So the maxima are ⌈10/2⌉ = 5, ⌈11/2⌉ = 6, 6, 7, 7, 8, …, ⌈29/2⌉ = 15.

Pair consecutive rows:
(10,11): 5+6 = 11, (12,13): 13, (14,15): 15, (16,17): 17, (18,19): 19, (20,21): 21, (22,23): 23, (24,25): 25, (26,27): 27, (28,29): 29.

Sum = 11+13+15+…+29 = 10 terms averaging 20 = **200**.

(400 is the total number of seats — half of them can be used.)

ANSWER 6: C

# Problem 7:
Equally spaced integers: aₙ = a₁ + (n−1)d.

From a₁₅ = a₁ + 14d with 241 ≤ a₁ + 14d ≤ 250 and 1 ≤ a₁ ≤ 10:
14d ≥ 241 − 10 = 231 and 14d ≤ 250 − 1 = 249, so 16.5 ≤ d ≤ 17.8 → d = 17.

From a₂ = a₁ + 17 with 13 ≤ a₁ + 17 ≤ 20: a₁ ≤ 3. Combined with a₁ + 238 ≥ 241 → a₁ ≥ 3. So a₁ = 3.

Check: a₁₅ = 3 + 14(17) = 241 ✓ (in range), a₂ = 20 ✓.

a₁₄ = 3 + 13(17) = 3 + 221 = 224. Digit sum = 2 + 2 + 4 = **8**.

ANSWER 7: A

# Problem 8:
Total = 1+2+⋯+9 = 45, so each group sums to 45/3 = 15.

Consider the group containing 9: the other two cards sum to 6, so they are {1,5} or {2,4} ({3,3} impossible).

**Case {9,1,5}:** remaining {2,3,4,6,7,8}. The group with 8 needs two cards summing to 7: only {3,4} (since {1,6} uses the used 1). Then the last group is {2,6,7}, which sums to 15 ✓.
→ {1,5,9}, {3,4,8}, {2,6,7}

**Case {9,2,4}:** remaining {1,3,5,6,7,8}. The group with 8 needs two summing to 7: only {1,6}. Last group {3,5,7} sums to 15 ✓.
→ {2,4,9}, {1,6,8}, {3,5,7}

Exactly **2** ways.

ANSWER 8: C

# Problem 9:
There are 36 equally likely outcomes. Count those with product > 10 (note product = 10 does **not** count):

- Die 1 = 1: none (max 6)
- Die 1 = 2: need b > 5 → b = 6 → 1 way
- Die 1 = 3: need b > 10/3 → b = 4,5,6 → 3 ways
- Die 1 = 4: need b > 2.5 → b = 3,4,5,6 → 4 ways
- Die 1 = 5: need b > 2 → b = 3,4,5,6 → 4 ways
- Die 1 = 6: need b > 5/3 → b = 2,3,4,5,6 → 5 ways

Total = 0+1+3+4+4+5 = 17, so probability = **17/36**.

(1/2 = 18/36 would result from mistakenly including product = 10.)

ANSWER 9: B

# Problem 10:
Label Chris as the person who takes piece #1. With n students, the pieces Chris takes are #1, #(n+1), #(2n+1), … i.e. pieces numbered 1 + kn.

For Chris to take piece #100 we need 100 = 1 + kn, i.e. n divides 99.

Divisors of 99: 1, 3, 9, 11, 33, 99. Among the choices, only **11** works (99 = 9 × 11, so Chris takes pieces 1, 12, 23, …, 100).

Checking others: 99/10, 99/19, 99/20, 99/25 are not integers.

ANSWER 10: B

Problem 1:

We need find the one number among all permutations of \(2,4,5,7\) that is a multiple of another such number.

Any such four-digit number is between \(2457\) and \(7542\). Therefore, if a larger one is an integer multiple of a smaller one, the multiplier can only be \(2\) or \(3\).

### Could the multiplier be 2?

Suppose \(N=2M\), where both \(N\) and \(M\) use the digits \(2,4,5,7\).

- If \(M\) ends in \(2\), then \(N\) ends in \(4\). Looking at the tens digit, doubling \(4,5,\) or \(7\) gives an invalid digit or repeats the \(4\), so this cannot work.
- If \(M\) ends in \(7\), then \(N\) ends in \(4\), with a carry of \(1\). The tens digit must be \(2\), producing \(5\). Thus \(M\) must be either \(4527\) or \(5427\):
  \[
  2(4527)=9054,\qquad 2(5427)=10854,
  \]
  neither of which is a valid permutation.

So no pair has one number twice the other.

### Could the multiplier be 3?

The smaller number must be at most
\[
\frac{7542}{3}=2514.
\]
The only permutations of \(2,4,5,7\) below \(2514\) are
\[
2457,\qquad 2475.
\]

Now calculate:
\[
3(2457)=7371,
\]
not a permutation of \(2,4,5,7\), but
\[
3(2475)=7425.
\]

Thus \(7425\) is the desired number. The other choices are not \(3\) times a valid smaller permutation.

ANSWER 1: D

Problem 2:

Let \(n=94\). We are multiplying

\[
\underbrace{99\cdots 99}_{94\text{ nines}}
\cdot
\underbrace{44\cdots 44}_{94\text{ fours}}.
\]

For a smaller example,

\[
999\cdot444=443556.
\]

In general, the product has:

- \(93\) digits equal to \(4\),
- then a \(3\),
- then \(93\) digits equal to \(5\),
- then a final \(6\).

So the sum of its digits is

\[
93\cdot 4+3+93\cdot 5+6.
\]

Compute:

\[
372+3+465+6=846.
\]

Choices such as \(945\) or \(954\) might result from incorrectly using 94 copies of both the 4s and 5s; there are only \(93\) of each in the middle pattern.

ANSWER 2: A

Problem 3:

A student earns \(5\) points for first, \(3\) for second, and \(1\) for third.

We seek the smallest score that guarantees the student has strictly more points than every other student.

A score of \(13\) must be earned as

\[
5+5+3.
\]

That means the student wins two races and finishes second in the other race.

Consider any rival student:

- In each race our student wins, the rival can earn at most \(3\) points.
- In the race our student finishes second, the rival can earn at most \(5\) points.

Thus a rival can earn at most

\[
3+3+5=11.
\]

Therefore, \(13\) points guarantees first overall.

Now check why smaller relevant scores do not guarantee this:

- A score of \(11\) can be \(5+5+1\). Another student could finish second in the first two races and win the third:
  \[
  3+3+5=11,
  \]
  causing a tie.
- A score of \(10\) can be \(5+5+0\). Another student can earn
  \[
  3+3+5=11,
  \]
  and exceed that score.

Thus \(13\) is the smallest score that guarantees more points than everyone else.

ANSWER 3: D

Problem 4:

Let

- \(c\) = number correct,
- \(u\) = number unanswered.

The score is

\[
5c+u,
\]
with
\[
c+u\le 20.
\]

Check each choice.

- \(90=5(18)+0\), so 90 is possible.
- \(91=5(18)+1\), so 91 is possible.
- \(92=5(18)+2\), so 92 is possible.
- \(95=5(19)+0\), so 95 is possible.
- For \(97\), the closest possibility is
  \[
  97=5(19)+2.
  \]
  But that would require \(19+2=21\) questions, and there are only 20 questions.

So 97 is not possible.

ANSWER 4: E

Problem 5:

Maria buys 4 disks for \$5, so her cost per disk is

\[
\frac{5}{4}=\$1.25.
\]

She sells 3 disks for \$5, so her selling price per disk is

\[
\frac{5}{3}.
\]

Her profit per disk is

\[
\frac{5}{3}-\frac{5}{4}
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \$100 profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

Choices such as 120 confuse the profit with the selling price; 240 is the number that gives a total profit of \$100.

ANSWER 5: D

Problem 6:

The rows have from 10 seats up through

\[
10+19=29
\]

seats.

In a row with \(n\) seats, the maximum number of students who can sit with no two adjacent is

\[
\left\lceil \frac n2\right\rceil.
\]

Pair consecutive rows:

\[
\begin{array}{c|c}
\text{Rows} & \text{Maximum students} \\ \hline
10,11 & 5+6=11\\
12,13 & 6+7=13\\
14,15 & 7+8=15\\
\vdots & \vdots\\
28,29 & 14+15=29
\end{array}
\]

These totals are

\[
11,13,15,\ldots,29.
\]

This is an arithmetic sequence of 10 terms, with average

\[
\frac{11+29}{2}=20.
\]

Therefore the total is

\[
10\cdot20=200.
\]

The tempting answer \(400\) would be the total number of seats, but students cannot sit next to each other.

ANSWER 6: C

Problem 7:

The numbers are equally spaced, so let the common difference be \(d\). Then

\[
a_{15}=a_1+14d
\]
and
\[
a_2=a_1+d.
\]

Subtracting gives

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

Using the given ranges,

\[
241-20\le a_{15}-a_2\le250-13.
\]

Thus

\[
221\le13d\le237.
\]

Dividing by 13 gives

\[
17\le d\le18.
\]

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

If \(d=18\), then

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

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

Thus \(d=17\). Then

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

Since \(241\le a_{15}\le250\),

\[
241\le a_1+238\le250,
\]
so

\[
3\le a_1\le12.
\]

Also,

\[
a_2=a_1+17
\]
must be at most 20, so \(a_1\le3\). Therefore,

\[
a_1=3.
\]

Now

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

The sum of its digits is

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

ANSWER 7: A

Problem 8:

The sum of all the cards is

\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 equal-sum groups, each group must sum to

\[
45\div3=15.
\]

The group containing 9 must have two other cards totaling

\[
15-9=6.
\]

The possible pairs are

\[
1+5=6,\qquad 2+4=6.
\]

So there are two cases.

### Case 1: \( \{1,5,9\} \)

The remaining cards are \(2,3,4,6,7,8\). They can be grouped as

\[
\{2,6,7\},\qquad \{3,4,8\}.
\]

This gives one division:

\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

### Case 2: \( \{2,4,9\} \)

The remaining cards are \(1,3,5,6,7,8\). They can be grouped as

\[
\{1,6,8\},\qquad \{3,