🔬 Model Lab

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co

Problem 1:
We are asked for the fraction of allowable 7-digit telephone numbers that begin with 9 and end with 0. Allowable means first digit is not 0 or 1.

Total possible: The first digit can be 2,3,4,5,6,7,8,9. That is 8 choices.
Each of the other 6 digits can be 0-9, i.e. 10 choices each.
So total = 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10^6.

Favorable: first digit is 9: 1 choice. Last digit is 0: 1 choice.
The middle 5 digits are free: 10^5 choices.
So favorable = 1 × 10^5 × 1 = 10^5.

Fraction = 10^5 / (8 × 10^6) = 1 / (8 × 10) = 1/80.

This is not 1/100, which would ignore the restriction on the first digit, nor 1/90, etc.
ANSWER 1: B

Problem 2:
We have implications:
Alan gets A ⇒ Beth gets A.
Beth gets A ⇒ Carlos gets A.
Carlos gets A ⇒ Diana gets A.
All three implications are true. Exactly two of the four get A's.

If Alan got an A, then by chaining, Beth, Carlos, Diana all get A's. That is 4 A's, contradicting exactly 2. So Alan did not get an A.

If Beth got an A, then Carlos must get an A, and then Diana must get an A. That is at least Beth, Carlos, Diana = 3 A's, contradicting exactly 2. So Beth did not get an A.

Thus Alan and Beth are not A-recipients. The two recipients must be the other two: Carlos and Diana.

Check consistency: Alan didn't get A, so "If Alan...then..." is vacuously true. Similarly Beth's statement is vacuously true. Carlos got A and Diana got A, so "If Carlos...then Diana..." is true. This fits exactly 2 A's.

So the pair cannot involve Alan or Beth.
ANSWER 2: C

Problem 3:
Add n to {3,6,9,10}. Mean of the 5 numbers = (3+6+9+10+n)/5 = (28+n)/5.
Median of the 5 numbers = middle number after sorting. It must be one of the numbers in the set.

Median must be 6, 9, or n.

Case median = 6: Then (28+n)/5 = 6, so 28+n=30, n=2.
Check: set {3,6,9,10,2} sorted is 2,3,6,9,10, median 6, mean 30/5=6. Works.

Case median = 9: Then (28+n)/5 = 9, so 28+n=45, n=17.
Check: set {3,6,9,10,17} sorted is 3,6,9,10,17, median 9, mean 45/5=9. Works.

Case median = n: Then (28+n)/5 = n, so 28+n=5n, 28=4n, n=7.
Check: set {3,6,9,10,7} sorted is 3,6,7,9,10, median 7, mean 35/5=7. Works.

No other median is possible. Indeed if n ≤6 the sorted list has 6 in the middle; if n>10 the middle is 9; if 6<n≤9 the middle is n, etc., all covered above.

Thus there are 3 possible values: 2,7,17.
ANSWER 3: C

Problem 4:
Need N, 100 ≤ N ≤ 999, with:
N ≡ 2 mod 6, N ≡ 5 mod 9, N ≡ 7 mod 11.

First combine mod 6 and mod 9. Write N = 9a+5.
Need 9a+5 ≡ 2 (mod 6). Since 9 ≡ 3 mod 6, 3a+5 ≡ 2 mod 6, so 3a ≡ 3 mod 6.
This happens iff a is odd. Indeed 3×even ≡0, 3×odd ≡3 mod 6.
So a = 2t+1, N = 9(2t+1)+5 = 18t+14.
So N ≡ 14 mod 18. This is the LCM condition.

Now N ≡ 7 mod 11: 18t+14 ≡ 7 mod 11.
18 ≡ 7 mod 11, 14 ≡ 3 mod 11, so 7t+3 ≡ 7, 7t ≡ 4 mod 11.
Inverse of 7 mod 11 is 8 since 7×8=56≡1. So t ≡ 4×8=32 ≡10 mod 11.
So t = 11s+10, N = 18(11s+10)+14 = 198s+194.

So N = 194 + 198s.
s=0 →194, s=1→392, s=2→590, s=3→788, s=4→986, s=5→1184 >999.
All five are three-digit and satisfy the remainders.

Thus there are 5 such integers.
ANSWER 4: E

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

Candace: let first-half baskets be x out of 12, second-half be y out of 18.
Steph higher each half:
x/12 < 15/20 = 0.75 ⇒ x < 9 ⇒ x ≤ 8.
y/18 < 10/10 = 1 ⇒ y < 18 ⇒ y ≤ 17.

Same overall percentage: (x+y)/(12+18) = 25/30.
So (x+y)/30 = 25/30 ⇒ x+y = 25.

With x ≤8, y ≤17, and x+y=25, we must have y =25−x ≥ 17. But y ≤17. Hence y=17 and x=8. Indeed 8/12≈66.7%<75% and 17/18≈94.4%<100%.

Difference second minus first = y−x = 17−8 = 9. This is Simpson's paradox.
ANSWER 5: C

Problem 6:
Let total team points be T.
Alexa scored T/4, Brittany scored 2T/7. For integer points, T must be divisible by 4 and 7, i.e. by 28. Write T=28k.

Then Alexa=7k, Brittany=8k. Chelsea=15. Let Others O = rest by 7 teammates.
O = T −7k −8k −15 = 13k−15.

Each of the 7 others scored at most 2, and at least 0, so 0 ≤ O ≤ 14.
Thus 0 ≤ 13k−15 ≤ 14.
Lower: 13k ≥15 ⇒ k ≥2 (k integer >1.15).
Upper: 13k ≤29 ⇒ k ≤2 (k integer <2.23).
Hence k=2.

Then O = 26−15 = 11. Check: T=56, Alexa=14, Brittany=16, Chelsea=15, others=11, sum 56. 11 points among 7 players with ≤2 each is possible, e.g. 2+2+2+2+2+1+0.
ANSWER 6: B

Problem 7:
Number N is four-digit, last two digits 99. So N = ...99.
Also N = m²−1 for some m. So N+1 = m² ends in 00. Hence m is a multiple of 10. Write m=10t.
Then N = 100t²−1.
For 1000 ≤ N ≤ 9999: 1001 ≤100t² ≤10000 ⇒ t=4,5,6,7,8,9,10.
This gives: 1599,2499,3599,4899,6399,8099,9999.

Also N = (m−1)(m+1) = (10t−1)(10t+1), with the two factors differing by 2 and coprime.

If N is the product of exactly two primes, say p×q, then with 1 < A < B, A×B=N, both A and B must be prime; if either were composite N would have at least three prime factors counting multiplicity. So we need both 10t−1 and 10t+1 prime.

Check:
t=4: 39 composite. No.
t=5: 49 composite. No.
t=6: 59 and 61 both prime. Yes: 3599=59×61.
t=7: 69 composite. No.
t=8: 79 prime but 81 composite. No.
t=9: 89 prime but 91=7×13. No.
t=10: 99 composite. No.

Only one number works. Note t=8 gives 79×3⁴, which has two distinct primes but four prime factors with multiplicity, so it is not a product of exactly two primes.
ANSWER 7: B

Problem 8:
Digits are strictly increasing left to right, so no repeats and order is forced. Leading zero impossible, so digits ≥1. Sum of squares =50, so no digit ≥8 since 8²=64>50.

Squares: 1,4,9,16,25,36,49.

Max digit 7: 7²=49, remainder 1 ⇒ digit 1. Set {1,7} → 17.

Max digit 6: 6²=36, remainder 14 from {1,2,3,4,5}. Squares 1,4,9,16,25. Only 9+4+1=14 works ⇒ {1,2,3,6} →1236.

Max digit 5: 5²=25, remainder 25 from digits 1-4. Only 16+9=25 works ⇒ {3,4,5} →345.

Max digit ≤4: max sum 1+4+9+16=30<50 impossible.

So the only such integers are 17, 1236, 345. Numerically largest is the 4-digit 1236.

Product of its digits =1×2×3×6=36.
ANSWER 8: C

Problem 9:
For nonzero real u, u/|u| = 1 if u>0, −1 if u<0. Let X=a/|a|, Y=b/

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are asked to find the fraction of all valid 7-digit telephone numbers (which cannot begin with $0$ or $1$) that begin with $9$ and end with $0$.

**Step-by-step solution:**
1. **Total number of possible telephone numbers:**
   - The first digit can be any digit from $2$ through $9$, giving $8$ choices.
   - Each of the remaining $6$ digits can be any of the $10$ digits from $0$ through $9$.
   - Thus, the total number of telephone numbers is:
     $$8 \times 10^6$$

2. **Number of favorable telephone numbers:**
   - The first digit must be $9$ ($1$ choice).
   - The last digit must be $0$ ($1$ choice).
   - The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
   - Thus, the number of favorable telephone numbers is:
     $$1 \times 10^5 \times 1 = 10^5$$

3. **Compute the fraction:**
   $$\text{Fraction} = \frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 1: B

---

### Problem 2:
**Restatement:** We have four students (Alan, Beth, Carlos, Diana). The implications are:
- Alan $\implies$ Beth
- Beth $\implies$ Carlos
- Carlos $\implies$ Diana

All statements are true, but exactly two students received an A. We need to determine which two received an A.

**Step-by-step solution:**
1. If Alan received an A, then Beth must have received an A, which implies Carlos received an A, which implies Diana received an A. That would mean all $4$ students received an A, contradicting the fact that only two did. Thus, **Alan did not receive an A**.
2. If Beth received an A, then Carlos and Diana must have also received an A. That would mean at least $3$ students received an A, again a contradiction. Thus, **Beth did not receive an A**.
3. Since neither Alan nor Beth received an A, and exactly two students received an A, those two must be **Carlos and Diana**.
4. Let's verify truth values:
   - Alan's statement: False $\implies$ False (True)
   - Beth's statement: False $\implies$ True (True)
   - Carlos's statement: True $\implies$ True (True)

All statements hold, and exactly Carlos and Diana received an A.

ANSWER 2: C

---

### Problem 3:
**Restatement:** A fifth number $n$ is added to the set $\{3, 6, 9, 10\}$ such that the mean of the resulting 5-number set equals its median. We want to find the number of possible values of $n$.

**Step-by-step solution:**
1. The mean of the 5 numbers is:
   $$\text{Mean} = \frac{3 + 6 + 9 + 10 + n}{5} = \frac{28 + n}{5}$$

2. The median depends on the value of $n$ relative to the sorted set $\{3, 6, 9, 10\}$:
   - **Case 1: $n \le 6$**
     The sorted order has $6$ in the middle, so $\text{Median} = 6$.
     $$\frac{28 + n}{5} = 6 \implies 28 + n = 30 \implies n = 2$$
     Since $2 \le 6$, $n = 2$ is a valid solution.

   - **Case 2: $6 \le n \le 9$**
     The sorted order has $n$ in the middle, so $\text{Median} = n$.
     $$\frac{28 + n}{5} = n \implies 28 + n = 5n \implies 4n = 28 \implies n = 7$$
     Since $6 \le 7 \le 9$, $n = 7$ is a valid solution.

   - **Case 3: $n \ge 9$**
     The sorted order has $9$ in the middle, so $\text{Median} = 9$.
     $$\frac{28 + n}{5} = 9 \implies 28 + n = 45 \implies n = 17$$
     Since $17 \ge 9$, $n = 17$ is a valid solution.

3. There are $3$ possible values of $n$: $\{2, 7, 17\}$.

ANSWER 3: C

---

### Problem 4:
**Restatement:** Find how many positive three-digit integers $N$ ($100 \le N \le 999$) satisfy:
$$N \equiv 2 \pmod 6, \quad N \equiv 5 \pmod 9, \quad N \equiv 7 \pmod{11}$$

**Step-by-step solution:**
1. Combine the first two congruences:
   - $N \equiv 5 \pmod 9 \implies N = 9k + 5$.
   - $N \equiv 2 \pmod 6 \implies N$ is even.
   - For $9k + 5$ to be even, $9k$ must be odd, which means $k$ is odd.
   - Let $k = 2m + 1$:
     $$N = 9(2m + 1) + 5 = 18m + 14$$
   - Thus, $N \equiv 14 \pmod{18}$.

2. Now combine with $N \equiv 7 \pmod{11}$:
   $$18m + 14 \equiv 7 \pmod{11}$$
   $$7m + 3 \equiv 7 \pmod{11} \implies 7m \equiv 4 \pmod{11}$$
   Multiply both sides by $8$ (since $7 \times 8 = 56 \equiv 1 \pmod{11}$):
   $$m \equiv 32 \equiv 10 \pmod{11}$$
   So $m = 11j + 10$ for some integer $j$.

3. Substitute back to find $N$:
   $$N = 18(11j + 10) + 14 = 198j + 180 + 14 = 198j + 194$$

4. Find the values of $j$ such that $100 \le N \le 999$:
   - For $j = 0$: $N = 194$
   - For $j = 1$: $N = 392$
   - For $j = 2$: $N = 590$
   - For $j = 3$: $N = 788$
   - For $j = 4$: $N = 986$
   - For $j = 5$: $N = 1184 > 999$

There are $5$ such three-digit integers.

ANSWER 4: E

---

### Problem 5:
**Restatement:** Steph scored $15/20$ in the first half and $10/10$ in the second half. Candace attempted $12$ in the first half and $18$ in the second half. Steph had a strictly higher percentage than Candace in each half, but they had the same overall percentage. Find how many more baskets Candace scored in the second half than the first.

**Step-by-step solution:**
1. Steph's overall record:
   - Total attempts: $20 + 10 = 30$
   - Total baskets: $15 + 10 = 25$
   - Overall percentage: $\frac{25}{30} = \frac{5}{6}$

2. Candace's attempts total $12 + 18 = 30$. Since her overall percentage matches Steph's, her total baskets scored must be $25$.
   - Let $c_1$ be baskets in the first half, and $c_2$ in the second half.
   - $c_1 + c_2 = 25$.

3. Apply the conditions for each half:
   - First half: $\frac{c_1}{12} < \frac{15}{20} = \frac{3}{4} \implies c_1 < 9 \implies c_1 \le 8$.
   - Second half: $\frac{c_2}{18} < \frac{10}{10} = 1 \implies c_2 < 18 \implies c_2 \le 17$.

4. Since $c_1 \le 8$ and $c_2 \le 17$, the maximum possible sum is $8 + 17 = 25$.
   Since $c_1 + c_2 = 25$, the only integer solution is:
   $$c_1 = 8 \quad \text{and} \quad c_2 = 17$$

5. The difference is:
   $$c_2 - c_1 = 17 - 8 = 9$$

ANSWER 5: C

---

### Problem 6:
**Restatement:** In a game, $\frac{1}{4}$ of the team's points were scored by Alexa, $\frac{2}{7}$ by Brittany, and Chelsea scored $15$ points. None o

# Competition Solutions

**Problem 1:** We need the fraction of valid phone numbers that start with 9 and end with 0.

- **Total valid numbers:** The first digit can be 2–9 (8 choices); the other 6 digits are free (10 choices each).
$$8 \times 10^6 = 8{,}000{,}000$$
- **Numbers starting with 9, ending with 0:** First digit fixed (1 way), last digit fixed (1 way), middle 5 digits free:
$$1 \times 10^5 \times 1 = 100{,}000$$
- **Fraction:**
$$\frac{100{,}000}{8{,}000{,}000} = \frac{1}{80}$$

*Wrong choice check:* Using $10^7$ as the total (forgetting that 0 and 1 are banned as first digits) gives $\frac{1}{100}$, choice E — incorrect.

ANSWER 1: B

---

**Problem 2:** Statements: A→B, B→C, C→D, all true, and exactly two A's.

- **If Alan got an A:** then B gets an A, forcing C, forcing D — that's **four** A's. Contradiction. So Alan did **not** get an A.
- **If Beth got an A:** then C gets an A, forcing D — that's **three** A's. Contradiction. So Beth did **not** get an A.
- **If Carlos got an A:** then Diana gets an A — exactly **two** A's: Carlos and Diana. Alan's and Beth's statements are vacuously true (they didn't get A's). ✓

Only **Carlos and Diana** works. Choices A, B, D, E all include Alan or Beth, which we ruled out (choice B would force Diana to also get an A, giving three A's).

ANSWER 2: C

---

**Problem 3:** Sum of $\{3, 6, 9, 10\}$ is 28, so the mean of five numbers is $\frac{28+n}{5}$. The median depends on where $n$ falls:

- **$n \le 6$:** the middle value is 6. Need $\frac{28+n}{5}=6 \Rightarrow n=2$. ✓ (indeed $2 \le 6$)
- **$6 < n < 9$:** the middle value is $n$ itself. Need $\frac{28+n}{5}=n \Rightarrow 28=4n \Rightarrow n=7$. ✓
- **$n \ge 9$:** the middle value is 9. Need $\frac{28+n}{5}=9 \Rightarrow n=17$. ✓

Possible values: $n = 2, 7, 17$ — that's **3** values.

ANSWER 3: C

---

**Problem 4:** Let $N$ be such a number. Notice:
$$2 = 6-4, \qquad 5 = 9-4, \qquad 7 = 11-4$$
So $N+4$ is divisible by 6, 9, **and** 11, hence by $\text{lcm}(6,9,11)=198$. Thus $N = 198k - 4$.

Three-digit values:
$$k=1: 194,\quad k=2: 392,\quad k=3: 590,\quad k=4: 788,\quad k=5: 986$$
($k=6$ gives 1184, which is four digits.) Spot-check 194: $194 = 6(32)+2 = 9(21)+5 = 11(17)+7$. ✓

That's **5** numbers.

ANSWER 4: E

---

**Problem 5:** Steph's overall percentage: $\frac{15+10}{20+10} = \frac{25}{30}$. Candace also took $12+18=30$ attempts, so she made **25 baskets total**. Let $x$ = her first-half makes, $y$ = her second-half makes: $x + y = 25$.

Steph beat her in each half:
- First half: $\frac{15}{20} = 75\% > \frac{x}{12} \Rightarrow x < 9 \Rightarrow x \le 8$
- Second half: $\frac{10}{10} = 100\% > \frac{y}{18} \Rightarrow y < 18 \Rightarrow y \le 17$

From $x+y=25$: $x \ge 25-17 = 8$. So $x = 8$ and $y = 17$.

Difference: $17 - 8 = \mathbf{9}$. (Check: $8/12 \approx 67\% < 75\%$, $17/18 \approx 94\% < 100\%$, total $25/30$ ✓)

ANSWER 5: C

---

**Problem 6:** Let $T$ = total points. Then Alexa gets $\frac{T}{4}$, Brittany $\frac{2T}{7}$, Chelsea 15, and the other seven members total $X$, where $0 \le X \le 14$ (each scored at most 2, and there are 7 of them).

$$\frac{T}{4}+\frac{2T}{7}+15+X = T \implies X = \frac{13T}{28}-15$$

Since points are whole numbers, $\frac{T}{4}$ and $\frac{2T}{7}$ must be integers, so $28 \mid T$. From $0 \le X \le 14$:
$$15 \le \frac{13T}{28} \le 29 \implies 32.3 \le T \le 62.5$$
The only multiple of 28 in this range is $T = 56$ (28 is too small — points would exceed the total; 84 gives $X = 24 > 14$).

$$X = \frac{13 \cdot 56}{28} - 15 = 26 - 15 = \mathbf{11}$$
Check: $14 + 16 + 15 + 11 = 56$ ✓, and 11 is achievable with seven players each ≤ 2 (e.g., 2+2+2+2+2+1+0).

ANSWER 6: B

---

**Problem 7:** Let $N$ end in 99 with $N = k^2 - 1$. Then $N + 1 = k^2$ ends in 00, so $100 \mid k^2$, forcing $10 \mid k$. Write $k = 10m$:
$$N = 100m^2 - 1 = (10m-1)(10m+1)$$
Four digits requires $1000 \le 100m^2-1 \le 9999$, so $4 \le m \le 10$.

For $N$ to be a product of **exactly two primes**, both factors must be prime (neither can equal 1, and if either were composite, $N$ would have ≥ 3 prime factors):

| $m$ | $(10m-1,\ 10m+1)$ | Both prime? |
|---|---|---|
| 4 | 39, 41 | 39 = 3·13 ✗ |
| 5 | 49, 51 | ✗ |
| 6 | **59, 61** | ✓ → $N = 3599$ |
| 7 | 69, 71 | 69 = 3·23 ✗ |
| 8 | 79, 81 | 81 = 3⁴ ✗ |
| 9 | 89, 91 | 91 = 7·13 ✗ |
| 10 | 99, 101 | ✗ |

Only $3599 = 59 \times 61 = 60^2 - 1$: exactly **1** number.

ANSWER 7: B

---

**Problem 8:** Digits strictly increase left to right, so they're all distinct, and 0 cannot appear (it would have to be the leading digit). Since $8^2 = 64 > 50$, all digits are at most 7. Find distinct digits from $\{1,\dots,7\}$ with squares summing to 50:

- **Using 7** ($49$): need 1 more: $1^2 = 1$ → digits $\{1,7\}$ → **17**
- **Using 6, not 7** ($36$): need 14 from $\{1,\dots,5\}$: $1+4+9=14$ → digits $\{1,2,3,6\}$ → **1236**
- **Using 5, not 6, 7** ($25$): need 25 from $\{1,2,3,4\}$: $9+16=25$ → digits $\{3,4,5\}$ → **345**
- **Without 5, 6, 7:** max is $1+4+9+16 = 26 < 50$ — impossible.

Candidates: 17, 345, 1236. The **largest** is $1236$ (check: $1+4+9+36 = 50$ ✓, and $1<2<3<6$ ✓).

Product of digits: $1 \cdot 2 \cdot 3 \cdot 6 = \mathbf{36}$.

ANSWER 8: C

---

**Problem 9:** Note $\frac{a}{|a|} = \text{sign}(a)$, and $\frac{abc}{|abc|} = \text{sign}(a)\text{sign}(b)\text{sign}(c)$.

Since $a+b+c = 0$ with all nonzero, they cannot all have the same sign. There are two cases:

**Case 1: two positive, one negative** (say $a, b > 0$, $c < 0$):
$$1 + 1 + (-1) = 1, \quad abc < 0 \Rightarrow \frac{abc}{|abc|} = -1 \quad\Rightarrow\quad 1 + (-1) = 0$$

**Case 2: two negative, one positive** (say $a, b < 0$, $c > 0$):
$$-1 + (-1) + 1 = -1, \quad abc > 0 \Rightarrow \frac{abc}{|abc|} = +1 \quad\Rightarrow\quad -1 + 1 = 0$$

In every case the value is $\mathbf{0}$ — the only possible value.

ANSWER 9: A

---

**Problem 10:** Track the fraction of the original block eaten each turn (each person eats half of what remains):

| Turn | Pers

# Problem 1:
Total possible numbers: the first digit can be 2–9 (8 choices), the other six digits are free, giving 8 · 10⁶ numbers.

Numbers beginning with 9 and ending in 0: first digit fixed (9), last digit fixed (0), middle five digits free: 1 · 10⁵ · 1 = 10⁵.

Fraction = 10⁵ / (8 · 10⁶) = 1/80.

(1/90 or 1/100 would come from wrongly allowing 9 or 10 choices for the first digit; there are only 8.)

ANSWER 1: B

# Problem 2:
The true implications are: Alan A ⇒ Beth A ⇒ Carlos A ⇒ Diana A.

- If Alan got an A, then Beth, Carlos, and Diana all did — that's 4 A's, too many. So Alan got no A (eliminates A and D).
- If Beth got an A, then Carlos and Diana did too — that's 3 A's. So Beth got no A (eliminates B and E).
- So the A's must be Carlos and Diana: Carlos's A forces Diana's A, and Alan's and Beth's statements are vacuously true since they got no A. Exactly two A's ✓.

ANSWER 2: C

# Problem 3:
Mean = (3+6+9+10+n)/5 = (28+n)/5. Consider where n falls.

- n ≤ 6: sorted list is (n,3,6,9,10) or (3,n,6,9,10); median = 6, so 28+n = 30 → n = 2 ✓ (2 ≤ 6).
- 6 ≤ n ≤ 9: sorted (3,6,n,9,10); median = n, so 28+n = 5n → n = 7 ✓ (in range).
- n ≥ 9: sorted (3,6,9,n,10) or (3,6,9,10,n); median = 9, so 28+n = 45 → n = 17 ✓ (17 ≥ 9).

Values: n = 2, 7, 17 → three possible values.

ANSWER 3: C

# Problem 4:
Let N be the number.
- N ≡ 2 (mod 6) and N ≡ 5 (mod 9): mod 18, N ≡ 14 works (14 ÷ 6 → r 2; 14 ÷ 9 → r 5). So N = 18k + 14.
- Need 18k + 14 ≡ 7 (mod 11): 7k + 3 ≡ 7 → 7k ≡ 4 (mod 11). Since 7·8 = 56 ≡ 1, inverse of 7 is 8: k ≡ 32 ≡ 10 (mod 11).

So k = 11m + 10, giving N = 198m + 194.

Three-digit values: 194, 392, 590, 788, 986 (next is 1184, too big) → 5 numbers.

ANSWER 4: E

# Problem 5:
Steph: 15 + 10 = 25 baskets in 30 attempts = 5/6 (≈83.3%).
Candace: x of 12 in first half, y of 18 in second, total 30 attempts, same rate: (x+y)/30 = 5/6 → x + y = 25.

Constraints:
- First half: Steph made 15/20 = 75%, so x/12 < 3/4 → x < 9 → x ≤ 8.
- Second half: Steph made 10/10 = 100%, so y/18 < 1 → y ≤ 17.

From x + y = 25 with y ≤ 17 we need x ≥ 8, and x ≤ 8 forces x = 8, y = 17.

Difference: 17 − 8 = 9.

ANSWER 5: C

# Problem 6:
Let T be the team total. Then the other 7 players scored
T − T/4 − 2T/7 − 15 = T(28 − 7 − 8)/28 − 15 = 13T/28 − 15.

T must be a multiple of 28. Also the other 7 players scored at most 7·2 = 14 points, and at least 0:
0 ≤ 13T/28 − 15 ≤ 14 → 15 ≤ 13T/28 ≤ 29 → 32.3 ≤ T ≤ 62.5.

The only multiple of 28 in range is T = 56, giving 13·56/28 − 15 = 26 − 15 = 11 (and 11 ≤ 14, achievable, e.g., five players with 2 and one with 1).

ANSWER 6: B

# Problem 7:
The number is k² − 1 and ends in 99, so k² ends in 00, meaning k = 10m and the number is 100m² − 1 = (10m − 1)(10m + 1).

Four digits: 1000 ≤ 100m² − 1 ≤ 9999 → m = 4, 5, …, 10.

For exactly two prime factors, both 10m − 1 and 10m + 1 must be prime:
- m = 4: 39 = 3·13 ✗
- m = 5: 49 = 7² ✗
- m = 6: 59 and 61 both prime ✓ → 3599 = 59·61
- m = 7: 69 = 3·23 ✗
- m = 8: 81 = 3⁴ ✗
- m = 9: 91 = 7·13 ✗
- m = 10: 99 = 9·11 ✗

Exactly one number works.

ANSWER 7: B

# Problem 8:
Digits are strictly increasing (so distinct), squares summing to 50. Search by number of digits:
- Five digits minimum: 1+4+9+16+25 = 55 > 50, impossible. So at most 4 digits.
- Four digits: 1+4+9+36 = 50 ✓ → 1236. (Other 4-element sets fail.)
- Two/three digits: {1,7} → 17; {3,4,5} → 345.

The largest integer has the most digits: 1236. Product of digits = 1·2·3·6 = 36.

(Answer 60 would come from 345, but 1236 is larger.)

ANSWER 8: C

# Problem 9:
Each term x/|x| is +1 or −1. Since a + b + c = 0 with all nonzero, either two are positive and one negative, or two negative and one positive.

Case 1 (two positive, one negative): a/|a| + b/|b| + c/|c| = 1 + 1 − 1 = 1. Also abc = (+)(+)(−) < 0, so abc/|abc| = −1. Sum = 1 − 1 = 0.

Case 2 (two negative, one positive): sum of first three = −1 − 1 + 1 = −1. And abc = (−)(−)(+) > 0, so abc/|abc| = +1. Sum = −1 + 1 = 0.

Only value is 0.

ANSWER 9: A

# Problem 10:
The successive bites are 1/2, 1/4, 1/8, 1/16, … of the original block, taken in order Sarika, Dev, Rajiv, Sarika, …

Sarika eats bites 1, 4, 7, …:
1/2 + 1/16 + 1/128 + ⋯, a geometric series with first term 1/2 and ratio 1/8.

Sum = (1/2)/(1 − 1/8) = (1/2)(8/7) = 4/7.

(Note 4/7 + 2/7 + 1/7 = 1, matching Dev's 2/7 and Rajiv's 1/7.)

ANSWER 10: A

Problem 1:

We want the fraction of allowable 7-digit telephone numbers that begin with \(9\) and end with \(0\).

- A telephone number cannot begin with \(0\) or \(1\), so its first digit has \(8\) choices: \(2,3,4,5,6,7,8,9\).
- Each of the other six digits has \(10\) choices.

Thus the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with \(9\) and end with \(0\):

- The first digit is fixed as \(9\).
- The last digit is fixed as \(0\).
- The five middle digits can be anything.

So there are

\[
10^5
\]

such numbers.

Therefore the desired fraction is

\[
\frac{10^5}{8\cdot10^6}=\frac1{80}.
\]

Choices such as \(\frac1{100}\) would incorrectly treat all first digits as equally possible, including \(0\) and \(1\).

ANSWER 1: B

Problem 2:

The statements are:

- Alan: If Alan gets an A, then Beth gets an A.
- Beth: If Beth gets an A, then Carlos gets an A.
- Carlos: If Carlos gets an A, then Diana gets an A.

Only two students received A’s.

If Alan got an A, then Beth would get an A, then Carlos would get an A, then Diana would get an A. That would give all four students A’s, impossible.

If Beth got an A, then Carlos and Diana would also get A’s. That would give at least three A’s, impossible.

So Alan and Beth cannot have A’s.

If Carlos gets an A, then Diana must also get an A. This produces exactly two A’s:

\[
\text{Carlos and Diana}.
\]

This satisfies every implication:

- Alan did not get an A, so Alan’s statement is automatically true.
- Beth did not get an A, so Beth’s statement is automatically true.
- Carlos did get an A, and Diana did too, so Carlos’s statement is true.

ANSWER 2: C

Problem 3:

The original set is

\[
\{3,6,9,10\}.
\]

Its sum is

\[
3+6+9+10=28.
\]

After adding \(n\), the mean is

\[
\frac{28+n}{5}.
\]

We consider where \(n\) falls in the ordered list.

### Case 1: \(n\le 6\)

The middle number is \(6\), so the median is \(6\).

Set mean equal to median:

\[
\frac{28+n}{5}=6.
\]

\[
28+n=30
\]

\[
n=2.
\]

This works, since \(2\le6\).

### Case 2: \(6<n<9\)

The ordered list is

\[
3,6,n,9,10,
\]

so the median is \(n\).

\[
\frac{28+n}{5}=n.
\]

\[
28+n=5n
\]

\[
28=4n
\]

\[
n=7.
\]

This works, since \(6<7<9\).

### Case 3: \(n\ge 10\)

The median is \(9\).

\[
\frac{28+n}{5}=9.
\]

\[
28+n=45
\]

\[
n=17.
\]

This works, since \(17\ge10\).

So the possible values are

\[
2,\ 7,\ 17.
\]

There are \(3\) possible values.

ANSWER 3: C

Problem 4:

We need positive three-digit integers \(x\) such that

\[
x\equiv2\pmod6,
\]

\[
x\equiv5\pmod9,
\]

and

\[
x\equiv7\pmod{11}.
\]

First combine the conditions modulo \(6\) and \(9\).

Numbers that leave remainder \(2\) when divided by \(6\) include

\[
2,8,14,20,26,\ldots
\]

Of these, \(14\) leaves remainder \(5\) when divided by \(9\). The pattern repeats every

\[
\operatorname{lcm}(6,9)=18.
\]

Thus

\[
x=14+18k.
\]

Now require remainder \(7\) upon division by \(11\):

\[
14+18k\equiv7\pmod{11}.
\]

Since \(14\equiv3\pmod{11}\) and \(18\equiv7\pmod{11}\),

\[
3+7k\equiv7\pmod{11}.
\]

\[
7k\equiv4\pmod{11}.
\]

Since \(7\cdot8=56\equiv1\pmod{11}\), multiply by \(8\):

\[
k\equiv 32\equiv10\pmod{11}.
\]

So \(k=10\) gives the first solution:

\[
x=14+18(10)=194.
\]

The full pattern repeats every

\[
\operatorname{lcm}(6,9,11)=198.
\]

Therefore the three-digit solutions are

\[
194,\ 392,\ 590,\ 788,\ 986.
\]

There are \(5\) such integers.

ANSWER 4: E

Problem 5:

Steph made

\[
15+10=25
\]

baskets out of

\[
20+10=30
\]

attempts. Her overall percentage was

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

Candace also took

\[
12+18=30
\]

attempts. Since she had the same overall percentage as Steph, Candace must also have made

\[
\frac56\cdot30=25
\]

baskets total.

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

\[
25-x
\]

baskets in the second half.

Steph’s first-half percentage was

\[
\frac{15}{20}=\frac34.
\]

Candace’s first-half percentage was lower, so

\[
\frac{x}{12}<\frac34.
\]

\[
x<9.
\]

Thus

\[
x\le8.
\]

Steph made all \(10\) baskets in the second half, a \(100\%\) rate. Candace’s second-half percentage was lower:

\[
\frac{25-x}{18}<1.
\]

\[
25-x<18
\]

\[
x>7.
\]

So

\[
x=8.
\]

Candace made \(8\) baskets in the first half and

\[
25-8=17
\]

in the second half. The difference is

\[
17-8=9.
\]

ANSWER 5: C

Problem 6:

Let the team’s total number of points be \(T\).

Alexa scored

\[
\frac14T,
\]

and Brittany scored

\[
\frac27T.
\]

Chelsea scored \(15\) points. Let \(R\) be the number of points scored by the other \(7\) team members.

Then

\[
R=T-\frac14T-\frac27T-15.
\]

Using a denominator of \(28\),

\[
R=T-\frac7{28}T-\frac8{28}T-15.
\]

\[
R=\frac{13}{28}T-15.
\]

Because Alexa’s and Brittany’s scores are whole numbers, \(T\) must be divisible by both \(4\) and \(7\), hence by \(28\). Write

\[
T=28k.
\]

Then

\[
R=13k-15.
\]

The other \(7\) players each scored at most \(2\) points, so together they scored at most

\[
7\cdot2=14
\]

points. Therefore

\[
13k-15\le14.
\]

\[
13k\le29.
\]

\[
k\le2.
\]

Also \(k=1\) would give \(R=13-15=-2\), which is impossible. Thus \(k=2\).

So

\[
R=13(2)-15=11.
\]

ANSWER 6: B

Problem 7:

The number ends in \(99\) and is \(1\) less than a perfect square. So it has the form

\[
m^2-1=(m-1)(m+1).
\]

Since \(m^2-1\) ends in \(99\), \(m^2\) ends in \(00\). Therefore \(m\) must end in \(0\).

The relevant values of \(m\) are

\[
40,50,60,70,80,90,100.
\]

We check

\[
m^2-1=(m-1)(m+1).
\]

For the number to be the product of exactly two primes, both factors \(m-1\) and \(m+1\) must be prime.

\[
40^2-1=39\cdot41
\]

but \(39\) is composite.

\[
50^2-1=49\cdot51
\]

and both factors are composite.

\[
60^2-1=59\cdot61.
\]

Both \(59\) and \(61\) are prime, so

\[
3599=59\cdot61
\]

works.

\[
70^2-1=69\cdot71
\]

but \(69\) is composite.

\[
80^2-1=79\cdot81
\]

but \(81\) is composite.

\[
90^2-1=89\cdot91
\]

but \(91=7\cdot13\) is composite.

\[
100^2-1=99\cdot101
\]

but \(99\) is co