🔬 Model Lab

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri

Problem 1:
We are asked how many disks give $100 profit when buying at 4 for $5 and selling at 3 for $5.

Cost of 1 disk = $5/4. Selling price of 1 disk = $5/3.
Profit on 1 disk = 5/3 - 5/4 = 20/12 - 15/12 = 5/12 dollars.

To avoid fractions, take 12 disks: cost = 3×5 = $15, revenue = 4×5 = $20, profit = $5 on 12 disks.

To make $100 need 100/5 = 20 such batches: 20×12 = 240 disks.

100 would assume $1 profit per disk, 120 would assume $5 profit per 6 disks, both too high a profit rate.
ANSWER 1: D

Problem 2:
We need a 5-digit number PQRST using 1,2,3,4,5 once each, with PQR divisible by 4, QRS divisible by 5, RST divisible by 3.

QRS is a 3-digit number divisible by 5, using digits 1-5. It must end in 0 or 5. 0 is unavailable, so S = 5.

Remaining digits P,Q,R,T are 1,2,3,4 in some order.

PQR divisible by 4 means its last two digits QR form a number divisible by 4. Possible 2-digit numbers from 1,2,3,4:
12,13,14,21,23,24,31,32,34,41,42,43.
Divisible by 4: 12, 24, 32 only. So QR is 12, 24, or 32.

RST = R 5 T divisible by 3 means sum R+5+T divisible by 3, i.e. R+T ≡ 1 mod 3.

Case QR=12: Q=1,R=2. Then P,T are 3,4. Need 2+T ≡1 mod 3, i.e. T≡2 mod 3. But 3≡0,4≡1. Impossible.

Case QR=24: Q=2,R=4. Then P,T are 1,3. Need 4+T≡1, i.e. T≡0 mod 3. So T=3, P=1. This gives 12453. Check:124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.

Case QR=32: Q=3,R=2. Then P,T are 1,4. Need 2+T≡1, i.e. T≡2 mod 3. But 1≡1,4≡1. Impossible.

Hence only P=1 works.
ANSWER 2: A

Problem 3:
600 students are split into 3 equal groups of 200. What is probability Al, Bob, Carol end in the same group?

Approximate by independent assignment with probability 1/3 each.

Fix Al's group. Bob is in same group with probability about 1/3. Carol is in same group with probability about 1/3. So all three same ≈ 1/3 × 1/3 = 1/9.

More exactly with equal sizes: after Al placed, 199 places remain in his group out of 599, then 198 out of 598, product ≈1/9.

1/27 is the probability they are all in a *specified* group, e.g. group 1. There are 3 groups, so multiply by 3 to get 1/9.
ANSWER 3: B

Problem 4:
Count n with n^3 between 2^8+1 and 2^18+1 inclusive.

Lower: 2^8+1 = 257. 6^3=216, 7^3=343. So cube root >6, smallest possible n is 7.

Upper: 2^18+1 = 262144+1 = 262145. Note 2^18 = (2^6)^3 = 64^3 = 262144. So upper is 64^3+1. 65^3=274625 is too big. So largest n with n^3 ≤ upper is 64.

Hence n = 7,8,...,64 inclusive. Count = 64-7+1 = 58.

Endpoints themselves are not cubes, so inclusive vs exclusive does not matter. 57 would omit one end incorrectly.
ANSWER 4: E

Problem 5:
Bus needs 2 min to drive one stop, 1 min wait at each stop. Zia walks 5 min per stop. Both start moving at time 0, bus 3 stops behind Zia.

Let Zia start at stop 0, bus start at stop -3 departing immediately.

Bus schedule:
leave -3 at 0, arrive -2 at 2; wait 2-3, arrive -1 at 5; wait 5-6, arrive 0 at 8; wait 8-9, arrive 1 at 11; wait 11-12, arrive 2 at 14; wait 14-15, arrive 3 at 17.

Zia schedule:
leave 0 at 0, arrive 1 at 5. Previous stop is 0. Bus at time 5 is just arriving at -1, not at or past 0, so she walks on.
leave 1 at 5, arrive 2 at 10. Previous stop is 1. Bus at time 10 is driving 0→1, not yet at 1, so she walks on.
leave 2 at 10, arrive 3 at 15. Previous stop is 2. Bus at time 15 is leaving 2, i.e. is at / has left the previous stop, so she waits.

Bus arrives at 3 at 17, when she boards. Total 17 minutes.

Choices like 19,20,21 assume extra waiting or walking further.
ANSWER 5: A

Problem 6:
Let original price be P and percent be r (as decimal). After increase: P(1+r). After decrease by same percent: P(1+r)(1-r) = P(1-r^2).

This equals 0.84 P, so 1-r^2 = 0.84, r^2 = 0.16, r = 0.4 = 40%.

16% would be the total loss, not the rate; it ignores compounding.
ANSWER 6: E

Problem 7:
Keiko tosses 1 coin: P(0 heads)=1/2, P(1 head)=1/2.
Ephraim tosses 2 coins: P(0)=1/4, P(1)=1/2, P(2)=1/4.

Same number of heads can only be 0 or 1, since Keiko cannot get 2.

P(both 0) = 1/2 × 1/4 = 1/8.
P(both 1) = 1/2 × 1/2 = 1/4 = 2/8.
Sum = 3/8.

1/4 omits one case, 1/2 assumes symmetry incorrectly.
ANSWER 7: B

Problem 8:
3 fish = 2 loaves. 1 loaf = 4 bags of rice.

So 2 loaves = 2×4 = 8 bags. Hence 3 fish = 8 bags. 1 fish = 8/3 = 2 2/3 bags.

Inverting the ratio gives 3/8, which is bags-to-fish, not fish-to-bags.
ANSWER 8: D

Problem 9:
Let initially green = 3k, yellow = k. Total = 4k.

3 green leave shade, 5 yellow enter shade: net green +2, net yellow -2. Total unchanged.

Now green = 3k+2, yellow = k-2, ratio 4:1:
(3k+2)/(k-2) = 4
3k+2 = 4k-8
k = 10.

Now green = 32, yellow = 8. Difference = 32-8 = 24.

Total is 40 throughout, consistent.
ANSWER 9: E

Problem 10:
We need a 4-digit number ending in 99, equal to m^2-1, and equal to product of exactly two primes.

Number ≡99 mod 100, so m^2-1≡99, m^2≡0 mod 100, so m is multiple of 10. Let m=10k.

Number = 100k^2-1, 1000 to 9999 gives k=4,...,10:
1599,2499,3599,4899,6399,8099,9999.
In factored form (10k-1)(10k+1).

For m^2-1 = (m-1)(m+1) to be product of exactly two primes, with both factors >1, both factors must be prime.

k=4: 39×41, 39 composite. Fail.
k=5: 49×51, composite. Fail.
k=6: 59×61, both prime. 3599 qualifies.
k=7: 69×71, 69 composite. Fail.
k=8: 79×81, 81 composite. Fail.
k=9: 89×91, 91=7×13. Fail.
k=10: 99×101, 99 composite. Fail.

Only 1 number qualifies. Note 6399 = 3^4×79 has 5 prime factors with multiplicity, so it does not count; exactly two distinct primes is not the meaning.
ANSWER 10: B

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

---

### Problem 1:
**Restatement & Setup:**  
Maria purchases computer disks at a rate of $4$ disks for $\$5$, which means the cost per disk is:
$$\text{Cost per disk} = \frac{\$5}{4} = \$1.25$$

She sells the disks at a rate of $3$ disks for $\$5$, so the revenue per disk is:
$$\text{Selling price per disk} = \frac{\$5}{3}$$

**Computation:**  
The profit per disk is:
$$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}$$

To make a total profit of $\$100$, the number of disks $N$ she must sell is:
$$N = \frac{100}{\frac{5}{12}} = 100 \times \frac{12}{5} = 20 \times 12 = 240$$

ANSWER 1: D

---

### Problem 2:
**Restatement & Setup:**  
The digits $1, 2, 3, 4, 5$ are each used once to form the five-digit number $PQRST$.
1. The three-digit number $QRS$ is divisible by $5$. Since the digits are non-zero, the units digit $S$ must be $5$.
2. The remaining digits for $P, Q, R, T$ are $\{1, 2, 3, 4\}$.
3. The three-digit number $PQR$ is divisible by $4$, which means the two-digit number $QR$ must be divisible by $4$.
   The possible pairs $(Q, R)$ from the digits $\{1, 2, 3, 4\}$ are:
   $$(1, 2), \quad (2, 4), \quad (3, 2)$$
4. The three-digit number $RST$ is divisible by $3$, which means the sum of its digits $R + S + T = R + 5 + T$ must be a multiple of $3$.

**Testing the cases:**
- **If $R = 2$:**  
  $R + 5 + T = 7 + T$ must be a multiple of $3$. Thus, $T$ could be $2, 5,$ or $8$. However, $2$ and $5$ are already used, and $8$ is not an available digit. Thus, no such $T$ exists, which eliminates $(Q, R) = (1, 2)$ and $(3, 2)$.

- **If $(Q, R) = (2, 4)$:**  
  $R + 5 + T = 4 + 5 + T = 9 + T$ must be divisible by $3$. The remaining digits for $\{P, T\}$ are $\{1, 3\}$.  
  For $9 + T$ to be divisible by $3$, $T$ must be $3$.  
  This leaves $P = 1$.

Checking the full number $PQRST = 12453$:
- $PQR = 124$ is divisible by $4$ ($124 = 4 \times 31$).
- $QRS = 245$ is divisible by $5$ ($245 = 5 \times 49$).
- $RST = 453$ is divisible by $3$ ($453 = 3 \times 151$).

Thus, $P = 1$.

ANSWER 2: A

---

### Problem 3:
**Restatement & Setup:**  
There are $600$ students divided equally into three groups of $200$. We want the probability that Al, Bob, and Carol are all assigned to the same lunch group.

**Computation:**  
- Al is assigned to one of the three lunch groups.
- The probability that Bob is assigned to the same group as Al is:
  $$\frac{199}{599} \approx \frac{1}{3}$$
- Given that Bob is in Al's group, the probability that Carol is also assigned to that same group is:
  $$\frac{198}{598} \approx \frac{1}{3}$$

Multiplying these independent selections gives:
$$\text{Probability} = \frac{199}{599} \times \frac{198}{598} \approx \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

ANSWER 3: B

---

### Problem 4:
**Restatement & Setup:**  
We are asked for the number of integer cubes $k^3$ such that:
$$2^8 + 1 \le k^3 \le 2^{18} + 1$$

**Computation:**  
1. Compute the lower bound:
   $$2^8 + 1 = 256 + 1 = 257$$
   Since $6^3 = 216 < 257$ and $7^3 = 343 \ge 257$, the smallest integer $k$ is $7$.

2. Compute the upper bound:
   $$2^{18} + 1 = (2^6)^3 + 1 = 64^3 + 1 = 262,145$$
   Since $64^3 \le 262,145 < 65^3$, the largest integer $k$ is $64$.

3. Count the number of integers from $7$ to $64$, inclusive:
   $$64 - 7 + 1 = 58$$

ANSWER 4: E

---

### Problem 5:
**Restatement & Setup:**  
- The bus takes $2$ minutes between stops and waits $1$ minute at each stop (a $3$-minute cycle per stop).
- Zia takes $5$ minutes to walk from one stop to the next.
- Let Stop $0$ be the bus's start point, so Zia starts at Stop $3$ at $t = 0$.
- Rule: When Zia arrives at a stop, she waits if the bus is at the previous stop or has already passed it; otherwise, she continues walking.

**Tracking step-by-step:**
- **$t = 0$ to $5$:** Zia walks from Stop $3$ to Stop $4$, arriving at $t = 5$.
  - Bus schedule: arrives at Stop $1$ at $t = 2$, departs at $t = 3$; arrives at Stop $2$ at $t = 5$.
  - At $t = 5$, the bus is at Stop $2$. The previous stop to Zia is Stop $3$. Since the bus has not reached Stop $3$, Zia keeps walking toward Stop $5$.
- **$t = 5$ to $10$:** Zia walks from Stop $4$ to Stop $5$, arriving at $t = 10$.
  - Bus schedule: departs Stop $2$ at $t = 6$, arrives at Stop $3$ at $t = 8$, departs Stop $3$ at $t = 9$, and is in transit to Stop $4$.
  - At $t = 10$, the bus is between Stop $3$ and Stop $4$. The previous stop to Zia is Stop $4$. The bus has not reached Stop $4$, so Zia keeps walking toward Stop $6$.
- **$t = 10$ to $15$:** Zia walks from Stop $5$ to Stop $6$, arriving at $t = 15$.
  - Bus schedule: arrives at Stop $4$ at $t = 11$, departs at $t = 12$; arrives at Stop $5$ at $t = 14$, waits until $t = 15$.
  - At $t = 15$, the bus is at Stop $5$ (the previous stop relative to Stop $6$).
  - Since the bus is at the previous stop, Zia waits at Stop $6$.

**Boarding Time:**  
The bus departs Stop $5$ at $t = 15$ and drives for $2$ minutes, arriving at Stop $6$ at:
$$t = 15 + 2 = 17\text{ minutes}$$
Zia boards the bus at $17$ minutes.

ANSWER 5: A

---

### Problem 6:
**Restatement & Setup:**  
Let the original price of the shirt be $P$, and let the price be increased by $x\%$ and then decreased by $x\%$.  
The new price after the increase is:
$$P\left(1 + \frac{x}{100}\right)$$
The price after the decrease is:
$$P\left(1 + \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) = P\left(1 - \left(\frac{x}{100}\right)^2\right)$$

**Computation:**  
We are given that the final price is $84\%$ of $P$:
$$1 - \left(\frac{x}{100}\right)^2 = 0.84$$
$$\left(\frac{x}{100}\right)^2 = 1 - 0.84 = 0.16$$
$$\frac{x}{100} = \sqrt{0.16} = 0.40 \implies x = 40$$

The price was increased and decreased by $40\%$.

ANSWER 6: E

---

### Problem 7:
**Restatement & Setup:**  
- Keiko tosses $1$ coin:
  - $0$ heads with probability $\frac{1}{2}$
  - $1$ head with probability $\frac{1}{2}$
- Ephraim tosses $2$ coins:
  

# Problem 1
Maria buys disks at 4 for $5 and sells them at 3 for $5.

**Cost per disk:** $5/4 = $1.25. **Selling price per disk:** $5/3 ≈ $1.67.

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

**Disks needed for $100 profit:** 100 ÷ (5/12) = 100 × 12/5 = **240**.

*Check:* 240 disks cost 240/4 × $5 = $300; selling them brings 240/3 × $5 = $400; profit = $100. ✓

Choices like 120 or 200 don't give $100: e.g., 120 disks would yield only 120 × 5/12 = $50.

ANSWER 1: D

# Problem 2
Digits 1–5 are used once each in PQRST.

- **QRS divisible by 5** → S must be 5 (0 isn't available). So P, Q, R, T use {1, 2, 3, 4}.
- **PQR divisible by 4** → the two-digit number QR must be divisible by 4. From digits {1,2,3,4}, distinct: candidates are 12, 24, 32 (only these are multiples of 4).
- **RST divisible by 3** → R + S + T = R + 5 + T must be divisible by 3.

Test each case:
- **(Q,R) = (1,2):** P,T from {3,4}. Need 2+5+T divisible by 3: 7+3=10, 7+4=11. Neither works. ✗
- **(Q,R) = (3,2):** P,T from {1,4}. 7+1=8, 7+4=11. Neither works. ✗
- **(Q,R) = (2,4):** P,T from {1,3}. Need 4+5+T divisible by 3: 9+3=12 ✓, so T=3, P=1. ✓

Number: 12453. Verify: 124 = 4×31 ✓; 245 ends in 5 ✓; 453 → 4+5+3 = 12 ✓. So P = 1.

ANSWER 2: A

# Problem 3
Each group has 600/3 = 200 students, so each friend has a 1/3 chance of landing in any given group.

- Al lands in *some* group (this is guaranteed — any group works).
- Bob must land in Al's group: probability 200/600 = 1/3.
- Carol must land in the same group: probability 1/3.

**P(all three together) = 1 × 1/3 × 1/3 = 1/9** (≈ 0.11).

*Tempting errors:* 1/27 wrongly forces a specific group for all three; 1/3 forgets that Carol must also match.

ANSWER 3: B

# Problem 4
Bounds: 2⁸ + 1 = 257 and 2¹⁸ + 1 = 262144 + 1 = 262145. We need n³ with 257 ≤ n³ ≤ 262145.

- **Smallest n:** 6³ = 216 < 257, but 7³ = 343 ≥ 257, so n starts at 7.
- **Largest n:** 64³ = 262144 ≤ 262145 (it counts!), and 65³ = 274625 > 262145, so n ends at 64.

**Count:** 64 − 7 + 1 = **58**.

*Trap:* 57 comes from wrongly excluding 64³ = 2¹⁸ = 262144, but 262144 ≤ 262145, so it's included.

ANSWER 4: E

# Problem 5
Number stops so Zia starts at stop 0; the bus starts at stop −3 and departs at t = 0.

**Bus schedule** (2 min drive + 1 min wait = 3 min per stop): arrives −2 at 2, leaves 3; arrives −1 at 5, leaves 6; **arrives 0 at 8, leaves 9; arrives 1 at 11, leaves 12; arrives 2 at 14, leaves 15; arrives 3 at 17.**
So the bus is *at* stop 0 during [8,9], stop 1 during [11,12], stop 2 during [14,15].

**Zia's arrivals** (5 min per stop), checking the *previous* stop each time:
- t = 5, stop 1: previous stop is 0. Bus doesn't reach stop 0 until t = 8, so it's not at/near it → **keep walking**.
- t = 10, stop 2: previous stop is 1. Bus doesn't reach stop 1 until t = 11 → **keep walking**.
- t = 15, stop 3: previous stop is 2. The bus is at stop 2 (there during [14,15], leaving at 15) → **Zia waits at stop 3**.

The bus arrives at stop 3 at **t = 17**, and Zia boards. (If she had walked on instead, she'd meet the bus at stop 4 at t = 20 — the tempting choice C — but the rule makes her wait.)

ANSWER 5: A

# Problem 6
Let the percent be x. Increasing by x% then decreasing by x% multiplies by (1 + x/100)(1 − x/100):

$$1 - \left(\frac{x}{100}\right)^2 = 0.84 \implies \left(\frac{x}{100}\right)^2 = 0.16 \implies \frac{x}{100} = 0.4 \implies x = 40.$$

*Check:* $100 → $100(1.40) = $140 → $140(0.60) = $84 = 84% of original. ✓

*Trap:* 16 is just 100 − 84, but percent changes don't cancel — the decrease acts on the larger price.

ANSWER 6: E

# Problem 7
- **Keiko (1 coin):** 0 heads with probability 1/2; 1 head with probability 1/2.
- **Ephraim (2 coins):** 0 heads with probability 1/4; 1 head with probability 1/2; 2 heads with probability 1/4.

They match only when both have 0 heads or both have 1 head:

$$P = \tfrac12\cdot\tfrac14 + \tfrac12\cdot\tfrac12 = \tfrac18 + \tfrac28 = \tfrac38.$$

*Trap:* 1/2 wrongly assumes Ephraim's heads count matches Keiko's half the time; the full distribution matters.

ANSWER 7: B

# Problem 8
- 3 fish = 2 loaves ⟹ **1 fish = 2/3 loaf**.
- 1 loaf = 4 bags of rice ⟹ 2/3 loaf = (2/3)(4) = **8/3 = 2⅔ bags**.

Check: 3 fish = 8 bags; 3 fish = 2 loaves = 8 bags ✓.

ANSWER 8: D

# Problem 9
Key insight: **frogs change color when they move** — shade makes them green, sun makes them yellow. So the 3 frogs leaving the shady (green) side become yellow, and the 5 frogs arriving at the shady side become green.

Let initial green = 3k, yellow = k. Then:
- New green = 3k − 3 (left) + 5 (arrived and turned green) = 3k + 2.
- New yellow = k − 5 + 3 (arrived and turned yellow) = k − 2.

New ratio 4 : 1 gives 3k + 2 = 4(k − 2) ⟹ 3k + 2 = 4k − 8 ⟹ **k = 10**.

Now: green = 32, yellow = 8 (ratio 4 : 1 ✓). **Difference = 32 − 8 = 24.**

*Note:* ignoring the color change gives 48 and 12 — still a 4:1 ratio — but their difference is 36, which isn't among the choices, confirming the color-change reading.

ANSWER 9: E

# Problem 10
The number ends in ...99. If N = ...99 and N + 1 is a perfect square, then N + 1 ends in 00, so its square root ends in 0: N + 1 = (10m)², giving N = 100m² − 1.

**Four-digit range:** 1000 ≤ 100m² − 1 ≤ 9999 ⟹ m ∈ {4, 5, 6, 7, 8, 9, 10} (m = 3 gives 899, too small).

**Factor:** N = 100m² − 1 = (10m − 1)(10m + 1). Both factors are greater than 1, so N is a product of exactly two primes **iff both factors are prime**:

| m | 10m−1 | 10m+1 | Prime pair? |
|---|-------|-------|-------------|
| 4 | 39 = 3·13 | 41 | No |
| 5 | 49 = 7² | 51 = 3·17 | No |
| 6 | 59 | 61 | **Yes** → N = 3599 |
| 7 | 69 = 3·23 | 71 | No |
| 8 | 79 | 81 = 3⁴ | No |
| 9 | 89 | 91 = 7·13 | No |
| 10 | 99 = 9·11 | 101 | No |

Only **3599 = 59 × 61** works: it ends in 99 ✓, and 3599 + 1 = 3600 = 60² ✓. So there is exactly **1** such number.

ANSWER 10: B

# Problem 1:
Cost: 4 disks for $5 → $5/4 = $1.25 per disk.
Selling: 3 disks for $5 → $5/3 per disk.

Profit per disk:
$$\frac{5}{3}-\frac{5}{4}=\frac{20-15}{12}=\frac{5}{12}\text{ dollars}$$

Number needed: $100 \div \frac{5}{12} = 100\cdot\frac{12}{5}=240$.

(Check: 240 disks cost $240\cdot1.25=\$300$, sell for $240\cdot\frac53=\$400$, profit $\$100$. ✓)
Tempting wrong answers come from using $\$5$ profit per group (giving 120 or 200), but the profit per *group of 12* disks is $\$5$, not per 3 or 4.

ANSWER 1: D

# Problem 2:
Digits 1–5 used once in $PQRST$.

**QRS divisible by 5:** last digit $S$ must be 0 or 5 → $S=5$.

**PQR divisible by 4:** then the two-digit ending $QR$ must be divisible by 4, so $R$ is even: $R\in\{2,4\}$.

**RST divisible by 3:** $R+S+T=R+5+T$ must be a multiple of 3.
- If $R=2$: $7+T\equiv0 \pmod 3 \Rightarrow T\equiv2\pmod 3 \Rightarrow T=2$ or $5$ — both already used. Impossible.
- If $R=4$: $9+T\equiv0\pmod 3\Rightarrow T$ is a multiple of 3 → $T=3$.

Remaining digits for $P,Q$: 1 and 2. Need $QR = Q4$ divisible by 4: $14$ is not, $24$ is → $Q=2$, so $P=1$.

Number: $12453$ (check: $124\div4=31$, $245\div5=49$, $453\div3=151$ ✓).

ANSWER 2: A

# Problem 3:
Fix Al's group. Each of Bob and Carol is independently placed in Al's group with probability $\frac13$.

$$P=\frac13\cdot\frac13=\frac19$$

($\frac1{27}$ would be the chance all three land in one *specific* named group; here the group is whichever Al gets.)

ANSWER 3: B

# Problem 4:
Range: from $2^8+1=257$ to $2^{18}+1=262145$, inclusive.

Smallest cube $\ge 257$: $6^3=216$ (too small), $7^3=343$ ✓ → start at $k=7$.
Largest cube $\le 262145$: $64^3=262144\le262145$ ✓, and $65^3=274625$ is too big → end at $k=64$.

Count: $64-7+1=58$.

(The trap is 57 — forgetting that $64^3 = 262144$ just squeezes under $2^{18}+1$.)

ANSWER 4: E

# Problem 5:
Label stops: Zia starts at stop 0; the bus starts 3 stops behind at stop $-3$.

**Bus schedule** (2 min travel, 1 min wait):
arrives $-2$ at 2 (leaves 3), arrives $-1$ at 5 (leaves 6), arrives 0 at 8 (leaves 9), arrives 1 at 11 (leaves 12), arrives 2 at 14 (leaves 15), arrives 3 at **17**.

**Zia** (5 min per stop), checking at each stop whether the bus has reached/left the previous stop:
- $t=0$, at stop 0: previous stop is $-1$; bus is still at $-3$ → walk.
- $t=5$, at stop 1: previous stop is 0; bus just arrived at $-1$ → hasn't reached 0 → walk.
- $t=10$, at stop 2: previous stop is 1; bus left stop 0 at 9 but reaches stop 1 only at 11 → hasn't reached stop 1 → walk.
- $t=15$, at stop 3: previous stop is 2; the bus **is at stop 2** (arrived 14, departs 15) → she waits.

Bus arrives at stop 3 at $t=17$, so she boards at 17 minutes.

ANSWER 5: A

# Problem 6:
Let the increase/decrease be $p$ (as a decimal). Price becomes
$$(1+p)(1-p)=1-p^2$$
of the original. Set equal to $0.84$:
$$1-p^2=0.84\Rightarrow p^2=0.16\Rightarrow p=0.4$$

So 40%. (16% is a trap: $0.16$ is $p^2$, not $p$.)

Check: $100\to140\to140(0.6)=84$ ✓

ANSWER 6: E

# Problem 7:
Keiko: 0 heads with prob $\frac12$, 1 head with prob $\frac12$.
Ephraim (2 pennies): 0 heads $\frac14$, 1 head $\frac24=\frac12$, 2 heads $\frac14$.

Match requires (0,0) or (1,1):
$$\frac12\cdot\frac14+\frac12\cdot\frac12=\frac18+\frac14=\frac38$$

ANSWER 7: B

# Problem 8:
$3$ fish $=2$ loaves → $1$ fish $=\frac23$ loaf.
$1$ loaf $=4$ bags of rice → $\frac23$ loaf $=\frac23\cdot4=\frac83=2\frac23$ bags.

ANSWER 8: D

# Problem 9:
Let green $=3x$, yellow $=x$.

After the moves: green $=3x-3+5=3x+2$, yellow $=x+3-5=x-2$.

New ratio $4:1$:
$$3x+2=4(x-2)\Rightarrow 3x+2=4x-8\Rightarrow x=10$$

So green $=32$, yellow $=8$; difference $=32-8=24$.

(24, not 20: the difference is asked for *now*, after the moves.)

ANSWER 9: E

# Problem 10:
The number is $n=k^2-1=(k-1)(k+1)$ and ends in "99", so $k^2$ ends in "00", meaning $k$ is a multiple of 10.

Four-digit range: $1000\le k^2-1\le 9999$ → $k=40,50,60,70,80,90,100$.

Now test whether $(k-1)(k+1)$ is a product of exactly two primes:
- $k=40$: $39\cdot41=3\cdot13\cdot41$ — three primes.
- $k=50$: $49\cdot51=7\cdot7\cdot3\cdot17$ — no.
- $k=60$: $59\cdot61$ — **both prime** ✓ ($n=3599$).
- $k=70$: $69\cdot71=3\cdot23\cdot71$ — no.
- $k=80$: $79\cdot81=79\cdot3^4$ — no.
- $k=90$: $89\cdot91=89\cdot7\cdot13$ — no.
- $k=100$: $99\cdot101=3^2\cdot11\cdot101$ — no.

Exactly one such number: 3599.

ANSWER 10: B

Problem 1:

Maria pays \( \$5\) for 4 disks, so her cost per disk is

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

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

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

Her profit per disk is

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

To earn \(\$100\), she must sell

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

disks.

For example, choice A, 100 disks, would produce only \(\frac{5}{12}(100)\), which is much less than \(\$100\). Choice E, 1200, is far too many.

\[
\boxed{240}
\]

ANSWER 1: D

Problem 2:

We need digits \(1,2,3,4,5\), each used once, such that:

- \(PQR\) is divisible by 4,
- \(QRS\) is divisible by 5,
- \(RST\) is divisible by 3.

Since \(QRS\) is divisible by 5, its last digit must be 5. Therefore,

\[
S=5.
\]

For \(PQR\) to be divisible by 4, its last two digits \(QR\) must be divisible by 4. Using digits \(1,2,3,4\), the possibilities are

\[
12,\quad 24,\quad 32.
\]

(The possible multiple \(52\) cannot be used because digit 5 is already \(S\).)

Now check each case using the rule that \(RST\) is divisible by 3.

### Case 1: \(QR=12\)

Then \(R=2\), \(S=5\), and \(T\) is either 3 or 4.

\[
2+5+3=10,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case fails.

### Case 2: \(QR=24\)

Then \(R=4\), \(S=5\), and the unused digits for \(P,T\) are 1 and 3.

For \(RST\) to be divisible by 3:

\[
4+5+T=9+T.
\]

We need \(T=3\), since

\[
4+5+3=12,
\]

which is divisible by 3. Therefore the remaining digit is

\[
P=1.
\]

### Case 3: \(QR=32\)

Then \(R=2\), \(S=5\), and \(T\) is 1 or 4.

\[
2+5+1=8,\qquad 2+5+4=11.
\]

Neither is divisible by 3, so this case also fails.

Thus,

\[
\boxed{P=1}.
\]

ANSWER 2: A

Problem 3:

There are 3 equal lunch groups, each with 200 students.

Al can be assigned to any group. Once Al’s group is chosen, Bob must be placed in that same group. Of the remaining 599 student spots, 199 are in Al’s group, so

\[
P(\text{Bob with Al})=\frac{199}{599}.
\]

Then Carol must also be placed in that group. Of the remaining 598 spots, 198 are in that group:

\[
P(\text{Carol with Al and Bob})=\frac{198}{598}.
\]

Therefore,

\[
\frac{199}{599}\cdot\frac{198}{598}\approx 0.110.
\]

Since

\[
\frac19\approx 0.111,
\]

the closest answer is \(\frac19\).

Choice A, \(\frac1{27}\), would incorrectly require all three students to be placed into one particular predetermined group. But the three friends may all be in any of the three groups.

\[
\boxed{\frac19}
\]

ANSWER 3: B

Problem 4:

We are asked for perfect cubes between

\[
2^8+1
\]

and

\[
2^{18}+1,
\]

inclusive.

Compute the endpoints:

\[
2^8+1=256+1=257,
\]

and

\[
2^{18}+1=262144+1=262145.
\]

Now find the first cube at least 257:

\[
6^3=216<257,
\]
\[
7^3=343>257.
\]

So the first cube is \(7^3\).

Find the last cube at most 262145:

\[
64^3=262144,
\]

which is allowed because it is less than \(262145\). But

\[
65^3=274625>262145.
\]

Thus the cubes are

\[
7^3,8^3,9^3,\ldots,64^3.
\]

The number of integers from 7 through 64 is

\[
64-7+1=58.
\]

A tempting error is to count from \(6^3\), but \(6^3=216\) is below the lower bound of 257.

\[
\boxed{58}
\]

ANSWER 4: E

Problem 5:

Label Zia’s starting stop as stop 0. The bus starts 3 stops behind her, at stop \(-3\).

The bus takes 2 minutes to travel between stops and waits 1 minute at each stop. Zia takes 5 minutes to walk between stops.

At time \(0\), the bus is not at Zia’s previous stop, so Zia starts walking.

- From time \(0\) to \(5\), Zia walks from stop 0 to stop 1.
- At time \(5\), the bus is still behind her, so she walks again.
- From time \(5\) to \(10\), Zia walks from stop 1 to stop 2.
- At time \(10\), the bus has not yet reached Zia’s previous stop, so she continues walking.
- From time \(10\) to \(15\), Zia walks from stop 2 to stop 3.

Meanwhile, the bus travels as follows:

\[
\begin{array}{c|c}
\text{Time} & \text{Bus location} \\ \hline
0 & \text{leaves stop }-3 \\
2 & \text{arrives at stop }-2 \\
5 & \text{arrives at stop }-1 \\
8 & \text{arrives at stop }0 \\
11 & \text{arrives at stop }1 \\
14 & \text{arrives at stop }2
\end{array}
\]

At time 15, Zia reaches stop 3. The bus is at the previous stop, stop 2, so Zia waits instead of walking further.

The bus leaves stop 2 at time 15 and takes 2 minutes to reach stop 3:

\[
15+2=17.
\]

Therefore, Zia boards the bus after 17 minutes.

\[
\boxed{17}
\]

ANSWER 5: A

Problem 6:

Let the original price be \(100\).

Suppose the price is increased by \(x\%\). Then the new price is

\[
100\left(1+\frac{x}{100}\right).
\]

Then it is decreased by the same \(x\%\), so the final price is

\[
100\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right).
\]

We are told the final price is \(84\%\) of the original price:

\[
\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=0.84.
\]

Using \((1+a)(1-a)=1-a^2\),

\[
1-\left(\frac{x}{100}\right)^2=0.84.
\]

So

\[
\left(\frac{x}{100}\right)^2=0.16.
\]

Thus

\[
\frac{x}{100}=0.4,
\]

and

\[
x=40.
\]

A common mistake is to think that increasing and decreasing by the same percent cancel. They do not, because the decrease is taken from the larger, increased price.

\[
\boxed{40\%}
\]

ANSWER 6: E

Problem 7:

Keiko tosses one penny. She gets either 0 heads or 1 head.

Ephraim tosses two pennies.

- If Keiko gets 0 heads, Ephraim must get 0 heads.
- If Keiko gets 1 head, Ephraim must get exactly 1 head.

The possible outcomes for Ephraim are:

\[
HH,\ HT,\ TH,\ TT.
\]

Thus,

\[
P(\text{Ephraim gets 0 heads})=\frac14,
\]

and

\[
P(\text{Ephraim gets 1 head})=\frac24=\frac12.
\]

Keiko gets 0 heads with probability \(\frac12\), and 1 head with probability \(\frac12\). Therefore,

\[
P(\text{same number of heads})
=
\frac12\cdot\frac14+\frac12\cdot\frac12.
\]

\[
=\frac18+\frac14
=\frac18+\frac28
=\frac38.
\]

\[
\boxed{\frac38}
\]

ANSWER 7: B

Problem 8:

We are told that

\[
3\text{ fish}=2\text{ loaves of bread}.
\]

Also,

\[
1\text{ loaf}=4\text{ bags of ri