🔬 Model Lab

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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

Problem 1:
We are asked for the rice value of 1 fish.
Given 3 fish = 2 loaves, and 1 loaf = 4 bags of rice.

Then 2 loaves = 2 x 4 bags = 8 bags of rice.
So 3 fish = 8 bags.

Divide by 3: 1 fish = 8/3 bags = 2 2/3 bags.

This rules out A,B,C which are <1, and E which is 3 1/3.
ANSWER 1: D

Problem 2:
Total seats = 4 rows x 3 = 12. 8 are taken, so 4 are empty.
The 4 empty seats are a random 4-subset of the 12.
The couple can sit together iff the empty set contains 2 seats
adjacent in the same row.

Total ways to choose 4 empties: C(12,4)=495.

Count complement: no adjacent empties in same row.
In one row with seats 1,2,3, forbidden to have both 1,2 or both 2,3.
Allowed: none, {1},{2},{3},{1,3}.
So 0 empties:1 way, 1 empty:3 ways, 2 empties:1 way ({1,3}), 3 empties:0 ways.

Over 4 rows we need total 4 with each 0-2:
- 1+1+1+1: 3^4 =81 ways.
- 2+1+1+0: choose 2-row:4, choose two 1-rows from 3:C(3,2)=3,
  1 way for the 2, 3^2 for the 1's: 4*3*9=108 ways.
- 2+2+0+0: choose 2 rows: C(4,2)=6 ways.

Total with no adjacent pair = 81+108+6=195.
So with at least one adjacent pair = 495-195=300.

Probability = 300/495 = 20/33.
ANSWER 2: C

Problem 3:
1200 students each take 5 classes.
Total student-seats = 1200 x 5 = 6000.

Each class holds 30 students, so number of class-sections
= 6000 / 30 = 200.

Each teacher teaches 4 classes, so number of teachers
= 200 / 4 = 50.
ANSWER 3: E

Problem 4:
A telephone number is 7 digits, first digit not 0 or 1.

First digit: 2,3,4,5,6,7,8,9 => 8 choices.
Each other digit: 10 choices.
Total = 8 x 10 x 10 x 10 x 10 x 10 x 10 = 8,000,000.

Favorable: begin with 9 and end with 0.
First digit fixed to 9: 1 choice.
Last digit fixed to 0: 1 choice.
Middle 5 digits free: 10^5 choices.
Favorable = 100,000.

Fraction = 100,000 / 8,000,000 = 1/80.
ANSWER 4: B

Problem 5:
Let total money be S, conserved by gifts.
Let initial be A0,J0,T0=36. So S = A0+J0+36.

Step1: Amy doubles Jan and Toy:
T1 = 72, J1 = 2J0, A1 = A0-J0-36.

Step2: Jan doubles Amy and Toy:
A2 = 2A1, T2 = 2T1 =144, J2 = J1-A1-T1.

Step3: Toy doubles Amy and Jan:
A3=2A2, J3=2J2, T3 = T2-A2-J2.

But T2+A2+J2 = S, so T3 = T2-A2-J2 = T2-(S-T2)=2T2-S.
Since T2=144, T3 = 288-S.

Given T3=36, so 288-S=36, S=252.

Note final Toy amount depends only on total, not distribution.
ANSWER 5: D

Problem 6:
5 groups of 5 from 1-25. Let group medians sorted be m1<=m2<=m3<=m4<=m5.
M = m3. Minimize M.

If M <= k, then at least 3 groups have median <=k.
A group with median <=k contains at least 3 numbers <=k
(the median and two smaller).
So at least 9 distinct numbers are <=k.
Since numbers are 1..25, this requires k >=9.
So M >=9.

Achievable: use numbers 1-9 three per group, fill with large numbers:
G1:1,2,3,24,25 median 3
G2:4,5,6,22,23 median 6
G3:7,8,9,20,21 median 9
G4:10,11,12,13,14 median 12
G5:15,16,17,18,19 median 17
Medians 3,6,9,12,17 have median 9.

So least possible is 9.
ANSWER 6: A

Problem 7:
Let radius = r = AB, since each passes through the other's center.
Put A at left, B at right. C is far left of A, D far right of B,
with C-A-B-D collinear, CA=AB=BD=r.

Let E be upper intersection. Then AE=AB=BE=r, so triangle ABE
is equilateral.

Set coordinates with r=1: A=(0,0), B=(1,0), C=(-1,0), D=(2,0),
E=(1/2, sqrt3/2).

Vectors from E:
EC = (-3/2, -sqrt3/2), ED = (3/2, -sqrt3/2).
|EC|^2 = 9/4+3/4=3, same for |ED|.
Dot = -9/4+3/4 = -3/2.
cos angle CED = (-3/2)/3 = -1/2.
So angle = 120 degrees.
ANSWER 7: C

Problem 8:
Statements are A_Alan -> A_Beth, A_Beth -> A_Carlos,
A_Carlos -> A_Diana, all true. Exactly two As.

If Alan had A, then Beth would, then Carlos, then Diana: 4 As,
impossible. So Alan has no A.

If Beth had A, then Carlos would, then Diana: at least 3 As,
impossible with Alan no A and only 2 total. So Beth has no A.

Since Alan,Beth have no A and exactly two have A, Carlos and Diana
must have A.

Check: Alan's and Beth's implications are vacuously true,
Carlos's is true-true, so true. Consistent.
ANSWER 8: C

Problem 9:
We need 1999^2000 mod 5.
1999 = 2000-1 ≡ -1 mod 5 (or 4 mod 5).
So 1999^2000 ≡ (-1)^2000 = 1 mod 5.
Remainder 1, not 0, so E out; not 4,3,2.
ANSWER 9: D

Problem 10:
Letters AJHSME length 6, all distinct, cycle period 6.
Digits 1989 length 4, cycle: 1989,9891,8919,9198,1989,... period 4.

Numbering starts with 1 = first shift:
1:JHSMEA 9891, 2:HSMEAJ 8919, etc.

We need first n>0 where letters return to AJHSME and digits to 1989.
Letters return iff n is multiple of 6. Digits return iff n is multiple of 4.
Need common multiple. lcm(6,4)=12.

So line 12 is the first repeat. Lines 6,18,24 are multiples of 6 but not 4; 10 is neither.
ANSWER 10: C

Problem 1:
We are asked to find how many bags of rice one fish is worth, given the trading rates between fish, bread, and rice.

First, set up the conversion equations:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

Using the second relation, $2$ loaves of bread can be traded for:
$$2 \times 4 = 8\text{ bags of rice}$$

Since $3$ fish can be traded for $2$ loaves of bread, it follows that:
$$3\text{ fish} = 8\text{ bags of rice}$$

Dividing both sides by $3$ to find the value of one fish:
$$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

Thus, one fish is worth $2\frac{2}{3}$ bags of rice.

ANSWER 1: D

---

Problem 2:
We are asked to find the probability that there will be $2$ adjacent empty seats in the same row after $8$ passengers have boarded a plane with $4$ rows of $3$ seats each.

The plane has $4 \times 3 = 12$ seats in total. Since $8$ passengers have boarded, there are $12 - 8 = 4$ empty seats remaining. 

The total number of ways to distribute the $4$ empty seats among the $12$ seats is:
$$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

Let us find the complementary probability, which is the number of ways to distribute the $4$ empty seats such that **no** row contains two adjacent empty seats.

In any row of $3$ seats (left, middle, right):
- $0$ empty seats: $1$ way (no adjacent empty seats).
- $1$ empty seat: $\binom{3}{1} = 3$ ways (no adjacent empty seats).
- $2$ empty seats: Only the left and right seats can be empty (not adjacent), which gives $1$ way.
- $3$ empty seats: All seats empty, which always contains adjacent empty seats ($0$ ways).

Now we count the ways to distribute the $4$ empty seats across the $4$ rows such that each row has at most $2$ empty seats and no row has adjacent empty seats:
1. **Two rows with 2 empty seats, two rows with 0 empty seats:**
   - Choose which 2 rows have 2 empty seats: $\binom{4}{2} = 6$ ways.
   - For each such row, there is only $1$ non-adjacent arrangement.
   - Total ways = $6 \times 1 \times 1 = 6$.

2. **One row with 2 empty seats, two rows with 1 empty seat, one row with 0 empty seats:**
   - Choose the row with 2 empty seats: $4$ choices (1 way to arrange seats).
   - Choose the row with 0 empty seats: $3$ choices.
   - The remaining 2 rows have 1 empty seat each: $3 \times 3 = 9$ ways to choose seats.
   - Total ways = $4 \times 3 \times 1 \times 9 = 108$.

3. **All four rows have 1 empty seat each:**
   - Each row has $3$ choices for its empty seat.
   - Total ways = $3^4 = 81$.

Adding these up, the total number of arrangements with **no** adjacent empty seats is:
$$6 + 108 + 81 = 195$$

Thus, the number of arrangements with at least $2$ adjacent empty seats in the same row is:
$$495 - 195 = 300$$

The probability is:
$$\frac{300}{495} = \frac{20}{33}$$

ANSWER 2: C

---

Problem 3:
We are given:
- King Middle School has $1200$ students.
- Each student takes $5$ classes per day.
- Each teacher teaches $4$ classes.
- Each class contains $30$ students and $1$ teacher.

We need to find the total number of teachers.

First, calculate the total number of student-class periods each day:
$$1200\text{ students} \times 5\text{ classes/student} = 6000\text{ student-class periods}$$

Since each class has $30$ students, the total number of classes offered is:
$$\frac{6000}{30} = 200\text{ classes}$$

Each teacher teaches $4$ classes, so the number of teachers needed is:
$$\frac{200}{4} = 50\text{ teachers}$$

ANSWER 3: E

---

Problem 4:
We are asked to find what fraction of valid 7-digit telephone numbers begin with $9$ and end with $0$, given that the first digit cannot be $0$ or $1$.

First, find the total number of possible telephone numbers:
- The first digit can be any digit from $2$ through $9$, so there are $8$ choices.
- The remaining $6$ digits can each be any of the $10$ digits from $0$ through $9$.
- Total valid telephone numbers = $8 \times 10^6$.

Next, count the telephone numbers that begin with $9$ and end with $0$:
- The first digit must be $9$ ($1$ choice).
- The last digit must be $0$ ($1$ choice).
- The middle $5$ digits can each be any of the $10$ digits ($10^5$ choices).
- Total favorable telephone numbers = $1 \times 10^5 \times 1 = 10^5$.

The desired fraction is:
$$\frac{10^5}{8 \times 10^6} = \frac{1}{8 \times 10} = \frac{1}{80}$$

ANSWER 4: B

---

Problem 5:
We are given that Amy, Jan, and Toy redistribute their money in three rounds:
1. Amy gives Jan and Toy enough to double their money.
2. Jan gives Amy and Toy enough to double their money.
3. Toy gives Amy and Jan enough to double their money.

Toy begins with $\$36$ and ends with $\$36$. We want to find the total amount of money, $S$, among the three friends. Note that $S$ remains constant throughout the trades.

Let Toy's amount at each stage be $T_0, T_1, T_2, T_3$:
- At the start, $T_0 = 36$.
- In Round 1, Amy doubles Toy's money:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72$$
- In Round 2, Jan doubles Toy's money:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144$$
- In Round 3, Toy doubles Amy's and Jan's money.
  Let Amy and Jan have amounts $A_2$ and $J_2$ before Round 3.
  Toy gives $A_2$ to Amy and $J_2$ to Jan, so Toy's final amount is:
  $$T_3 = T_2 - (A_2 + J_2)$$

Since the total money is $S$, we have $A_2 + J_2 + T_2 = S$, which means $A_2 + J_2 = S - T_2$. Substituting this into the equation for $T_3$:
$$T_3 = T_2 - (S - T_2) = 2T_2 - S$$

We are given that Toy ends with $\$36$, so $T_3 = 36$. Since $T_2 = 144$:
$$36 = 2(144) - S = 288 - S$$
$$S = 288 - 36 = 252$$

The total amount of money the three friends have is $\$252$.

ANSWER 5: D

---

Problem 6:
The integers $1$ through $25$ are partitioned into $5$ groups of $5$ numbers each. Let the medians of the five groups be ordered as:
$$m_1 \le m_2 \le m_3 \le m_4 \le m_5$$
Then $M = m_3$ is the median of the medians. We want the least possible value of $M$.

In each of the groups co

# Problem 1:
**Restate:** 3 fish = 2 loaves; 1 loaf = 4 bags of rice. Find the rice-value of 1 fish.

Set up: From the first trade, 1 fish = 2/3 loaf. From the second trade, 1 loaf = 4 bags of rice.

Compute: 1 fish = (2/3) × 4 = 8/3 = 2 2/3 bags of rice.

Check tempting wrong answers: 4 (choice E, 3 1/3, and others) come from mixing up the direction of the trades (e.g., treating 2 loaves = 3 fish as 3 loaves = 2 fish). The chain fish → loaves → rice must be followed in order.

ANSWER 1: D

# Problem 2:
**Restate:** 12 seats (4 rows × 3), 8 filled at random, so 4 empty seats chosen at random among 12. Find P(at least one adjacent pair of empty seats in the same row exists).

Complementary counting — count arrangements with **no** two empty seats adjacent in the same row. Each row of 3 seats allows:
- 0 empty seats: 1 way
- 1 empty seat: 3 ways
- 2 empty seats: only the pair {end, end} (seats 1 and 3), 1 way
- 3 empty seats: impossible (would contain adjacent seats)

Distribute 4 empty seats over 4 rows:
- **2+2+0+0:** choose the two rows with 2 empties: C(4,2) = 6; each such row has 1 way → 6
- **2+1+1+0:** 4 ways to pick the "2" row, 3 ways to pick the "0" row, 3 × 3 = 9 ways for the singles: 4 × 3 × 9 = 108
- **1+1+1+1:** 3⁴ = 81

No-adjacent total: 6 + 108 + 81 = 195. Total ways: C(12,4) = 495.

P(no adjacent pair) = 195/495 = 13/33, so P(adjacent pair exists) = 1 − 13/33 = **20/33**. (Wrong answers like 8/15 come from miscounting which pairs count as adjacent.)

ANSWER 2: C

# Problem 3:
**Restate:** 1200 students, 5 classes each; 30 students and 1 teacher per class; each teacher teaches 4 classes. Find the number of teachers.

Total student-class enrollments: 1200 × 5 = 6000.
Number of classes: 6000 ÷ 30 = 200.
Number of teachers: 200 ÷ 4 = **50**.

Common error: dividing 1200 by 30 and forgetting each student takes 5 classes (gives 40, a trap choice).

ANSWER 3: E

# Problem 4:
**Restate:** 7-digit numbers, first digit ≠ 0, 1. What fraction begin with 9 and end with 0?

Total possible numbers: first digit has 8 choices (2–9), other 6 digits free: 8 × 10⁶.

Favorable: first digit must be 9 (1 way), last digit must be 0 (1 way), middle 5 digits free: 10⁵.

Fraction = 10⁵ / (8 × 10⁶) = 1/80.

Traps: 1/81 comes from assuming all first digits equally likely including 0; 1/90 from wrongly allowing 9 first-digit choices with the last digit forced among 10.

ANSWER 4: B

# Problem 5:
**Restate:** Three rounds of doubling; Toy starts with $36 and ends with $36. Find the total T (which never changes, since money is only redistributed).

Track Toy's money:
- Start: $36
- After Amy doubles Toy: $72
- After Jan doubles Toy: $144

Now Toy doubles Amy's and Jan's money. Suppose together they hold S at that moment; Toy gives away S, ending with 144 − S = 36, so S = 108.

Total: T = Toy's 144 + 108 = **$252**.

Check: total is conserved throughout, so this is consistent. (Trap: 198 arises if you forget Toy must give *each* friend an amount equal to what they already hold.)

ANSWER 5: D

# Problem 6:
**Restate:** Split 1–25 into five groups of 5; M = median of the five group-medians. Minimize M.

**Lower bound:** If M is the median of the five medians, at least 3 groups have median ≤ M. Each such group contains at least 3 numbers ≤ M (two below its median plus the median). So we need at least 3 × 3 = 9 numbers ≤ M, forcing M ≥ 9.

**Construction achieving M = 9:** Use the numbers 1–9 as three "low triples":
- {1, 2, 3, 23, 24} → median 3
- {4, 5, 6, 22, 25} → median 6
- {7, 8, 9, 20, 21} → median 9
- {10, 11, 12, 13, 14} → median 12
- {15, 16, 17, 18, 19} → median 17

Medians: 3, 6, 9, 12, 17 → median of these = **9**. So the least possible M is 9.

ANSWER 6: A

# Problem 7:
**Restate:** Two congruent circles, each passing through the other's center; line AB extended meets them at C and D; E is an intersection point. Find ∠CED.

Set up coordinates: let the radius be 1, with A = (0,0), B = (1,0). Then C = (−1, 0), D = (2, 0), and E = (1/2, √3/2) (since ABE is equilateral with side 1).

Vectors from E:
- C − E = (−3/2, −√3/2)
- D − E = (3/2, −√3/2)

Dot product: (−3/2)(3/2) + (−√3/2)(−√3/2) = −9/4 + 3/4 = −3/2.
Magnitudes: each = √(9/4 + 3/4) = √3, product = 3.

cos ∠CED = (−3/2)/3 = −1/2 → ∠CED = **120°**. (Trap answers 90 or 135 come from misplacing C or D.)

ANSWER 7: C

# Problem 8:
**Restate:** True implications: Alan→Beth, Beth→Carlos, Carlos→Diana. Exactly two students got A's. Who?

If **Alan** got an A, then Beth did, then Carlos did, then Diana did — that's four A's. Contradiction, so Alan did **not** get an A.

If **Beth** got an A, then Carlos and Diana got A's — that's three A's (Beth, Carlos, Diana). Contradiction, so Beth did **not** get an A.

The two A's must come from {Carlos, Diana} — and indeed both can get A's while making only the Carlos→Diana implication relevant, which is satisfied. Exactly two A's: **Carlos and Diana**.

ANSWER 8: C

# Problem 9:
**Restate:** Find 1999²⁰⁰⁰ mod 5.

Since 1999 ≡ 4 ≡ −1 (mod 5),
1999²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).

So the remainder is **1**. (Trap: choosing −1 ≡ 4 by forgetting the exponent is even.)

ANSWER 9: D

# Problem 10:
**Restate:** The 6 letters cycle with one period, the 4 digits cycle with another; find the line where AJHSME 1989 first reappears in the numbered list.

Letters A, J, H, S, M, E: shifting one position each line, they return to AJHSME after **6** lines.
Digits 1, 9, 8, 9: 1989 → 9891 → 8919 → 9198 → 1989, so they return after **4** lines.

The full configuration AJHSME 1989 returns when both cycles align, i.e., after lcm(6, 4) = **12** steps of the cycle. With the list numbered as shown (the numbered entries being the successive cycled configurations, JHSMEA 9891 as entry 1), the configuration matching the start of the cycle, AJHSME 1989, first appears on line **12**.

Check the wrong choices: 6 works for letters alone (line 6 would show EAJHSM-type shifts, but the digits would be off), 4 for digits alone — neither aligns bot

# Problem 1:
We need the value of one fish in bags of rice.

- 3 fish = 2 loaves of bread
- 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice

Therefore 3 fish = 8 bags of rice, so

$$1 \text{ fish} = \frac{8}{3} = 2\tfrac23 \text{ bags of rice}.$$

Tempting wrong answers come from mixing up the direction (e.g. 3/8 is rice-per... reversed, and 3/4 or 1/2 come from misusing the 3-for-2 trade).

ANSWER 1: D

# Problem 2:
There are 12 seats; 8 are taken, so **4 seats are empty**. The couple can sit iff two empty seats are adjacent **within some row** (rows have seats 1‑2‑3, so adjacent pairs are (1,2) and (2,3)).

Total ways to choose which 4 seats are empty: $\binom{12}{4}=495$.

Count the "bad" cases (no two empty seats adjacent in a row). In a row of 3, allowed empty-sets:
- 0 empty: 1 way
- 1 empty: 3 ways
- 2 empty (must be seats 1 and 3): 1 way
- 3 empty: impossible

Distribute 4 empty seats among 4 rows with at most 2 per row:
- (2,2,0,0): choose the two rows with 2 empties: $\binom42=6$ ways → 6
- (2,1,1,0): pick the "2" row (4), the "0" row (3), each single-empty row 3 ways: $4\cdot3\cdot3\cdot3=108$
- (1,1,1,1): $3^4=81$

Bad total $=6+108+81=195$.

Probability the couple can sit $=1-\dfrac{195}{495}=\dfrac{300}{495}=\dfrac{20}{33}$.

ANSWER 2: C

# Problem 3:
Count class enrollments two ways.

Student-slots: $1200 \times 5 = 6000$.
Each class holds 30 students: number of classes $=6000/30 = 200$.
Each teacher teaches 4 classes (and each class has 1 teacher): teachers $=200/4 = 50$.

(Choice 40 would come from forgetting a factor; 30 from dividing 1200 by 40, etc.)

ANSWER 3: E

# Problem 4:
The first digit can be any of $2,3,\dots,9$ — that's 8 choices (not 10, since 0 and 1 are banned). The other 6 digits are free.

Total numbers: $8\cdot 10^6$.
Numbers starting with 9 and ending with 0: first digit fixed (1 way), last digit fixed (1 way), middle 5 free: $10^5$.

Fraction $=\dfrac{10^5}{8\cdot10^6}=\dfrac{1}{80}$.

(1/100 would be the answer if all 10 first digits were allowed.)

ANSWER 4: B

# Problem 5:
Let the total be $T$ (it never changes — money is only passed around). Track Toy:

- Start: Toy has 36.
- Amy doubles Jan's and Toy's amounts: Toy has $72$.
- Jan doubles Amy's and Toy's amounts: Toy has $144$.
- Toy doubles Amy's and Jan's amounts: Amy + Jan currently hold $T-144$, so Toy pays out $T-144$, leaving $144-(T-144)=288-T$.

Given the final amount is 36:
$$288 - T = 36 \implies T = 252.$$

ANSWER 5: D

# Problem 6:
Let the five medians be $m_1<m_2<M<m_4<m_5$ (all numbers distinct), where $M$ is the middle median.

**Lower bound:** Count numbers smaller than $M$:
- The group with median $m_1$: $m_1$ itself plus the 2 numbers below it → 3 numbers $<M$.
- The group with median $m_2$: likewise 3 numbers $<M$.
- The group with median $M$: 2 numbers below $M$.

So at least $3+3+2=8$ integers are less than $M$, forcing $M\ge 9$.

**$M=9$ is attainable:**
- $\{1,2,3,24,25\}$ → median 3
- $\{4,5,6,22,23\}$ → median 6
- $\{7,8,9,20,21\}$ → median 9
- $\{10,11,12,13,14\}$ → median 12
- $\{15,16,17,18,19\}$ → median 17

Medians: $3,6,9,12,17$; their median is $M=9$.

ANSWER 6: A

# Problem 7:
Let the radius be 1, $A=(0,0)$, $B=(1,0)$ (since each circle passes through the other's center, $AB=1$).

The line $AB$ meets circle $A$ at $(-1,0)$ and circle $B$ at $(2,0)$, so $C=(-1,0)$, $D=(2,0)$.

The circles intersect where $x^2+y^2=1$ and $(x-1)^2+y^2=1$: $x=\tfrac12,\ y=\tfrac{\sqrt3}{2}$, so $E=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$.

Vectors: $\vec{EC}=(-1.5,-\tfrac{\sqrt3}{2})$, $\vec{ED}=(1.5,-\tfrac{\sqrt3}{2})$.
Lengths: both $\sqrt{2.25+0.75}=\sqrt3$. Dot product: $-2.25+0.75=-1.5$.

$$\cos\angle CED=\frac{-1.5}{\sqrt3\cdot\sqrt3}=-\frac12 \implies \angle CED = 120^\circ.$$

(90° would be right only if $C,D$ were endpoints of one circle's diameter through $E$'s circle differently.)

ANSWER 7: C

# Problem 8:
The true statements chain: Alan A ⟹ Beth A ⟹ Carlos A ⟹ Diana A.

- If **Alan** got an A, then Beth, Carlos, Diana all did → 4 A's. Too many.
- If **Beth** got an A (Alan not), then Carlos and Diana did → 3 A's. Too many.
- If **Carlos** got an A (Alan, Beth not), then Diana did → exactly Carlos and Diana: 2 A's. ✔

So Carlos and Diana received the A's. (Note "Beth, Diana" is impossible since Beth's A forces Carlos's.)

ANSWER 8: C

# Problem 9:
We want $1999^{2000} \bmod 5$.

$1999 = 5\cdot 400 - 1 \equiv -1 \pmod 5$.

So $1999^{2000}\equiv(-1)^{2000}=1 \pmod 5$.

Remainder is 1 (an even exponent gives $+1$; an odd one would give $-1\equiv 4$).

ANSWER 9: D

# Problem 10:
The 6 letters return to AJHSME after every multiple of **6** cycles; the 4 digits return to 1989 after every multiple of **4** cycles (1989 → 9891 → 8919 → 9198 → 1989).

Both are simultaneously back to the original after $\operatorname{lcm}(6,4)=12$ cycles.

Since line $n$ shows the arrangement after $n$ cycles, "AJHSME 1989" first reappears on line **12**.

(6 fails: digits would read 8919; 10 fails for both.)

ANSWER 10: C

Problem 1:

We are asked for the number of bags of rice equal in value to 1 fish.

We know:

- 3 fish can be traded for 2 loaves of bread.
- 1 loaf of bread can be traded for 4 bags of rice.

So 2 loaves of bread are worth

\[
2\cdot 4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

Choice D is \(2\frac23\). Choices such as \(\frac34\) reverse the relationship rather than finding rice per fish.

ANSWER 1: D

Problem 2:

We need the probability that, after 8 passengers occupy 8 of the 12 seats, the 4 empty seats include two adjacent seats in the same row.

There are 4 rows of 3 seats each, for a total of

\[
4\cdot 3=12
\]

seats.

Since 8 seats are filled, there are 4 empty seats. Every set of 4 empty seats is equally likely.

The total number of ways to choose the 4 empty seats is

\[
\binom{12}{4}=495.
\]

It is easier to count the complement: arrangements in which there are **no** adjacent empty seats in any row.

For one row of 3 seats, the possible ways to choose empty seats with no adjacent empty seats are:

- 0 empty seats: \(1\) way
- 1 empty seat: \(3\) ways
- 2 empty seats: only the two end seats, so \(1\) way

We need 4 empty seats altogether. The possible row-distributions are:

1. \(2,2,0,0\)

\[
\binom42=6
\]

ways.

2. \(2,1,1,0\)

Choose the row with 2 empties, the row with 0 empties, and then choose the empty seat in each of the two 1-empty rows:

\[
4\cdot 3\cdot 3^2=108.
\]

3. \(1,1,1,1\)

Each row has 3 choices for its one empty seat:

\[
3^4=81.
\]

So the number with no adjacent empty seats is

\[
6+108+81=195.
\]

Therefore, the number with at least one adjacent pair of empty seats is

\[
495-195=300.
\]

Thus the desired probability is

\[
\frac{300}{495}=\frac{20}{33}.
\]

Choice C is \(\frac{20}{33}\). For example, \(\frac{32}{55}\) does not account correctly for cases with more than one possible adjacent pair.

ANSWER 2: C

Problem 3:

We are asked for the number of teachers.

There are 1200 students, and each takes 5 classes per day. Thus the total number of student-class enrollments is

\[
1200\cdot 5=6000.
\]

Each class has 30 students, so the number of class sections is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choice E is 50. A tempting mistake is to divide 1200 directly by 30, which counts classes for only one period rather than all 5 classes each student takes.

ANSWER 3: E

Problem 4:

Valid telephone numbers have 7 digits, but the first digit cannot be 0 or 1.

Thus the first digit has 8 choices:

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

The remaining 6 digits each have 10 choices, so the total number of possible telephone numbers is

\[
8\cdot 10^6.
\]

For a number to begin with 9 and end with 0:

- the first digit is fixed as 9;
- the last digit is fixed as 0;
- the 5 middle digits can be anything.

Therefore, there are

\[
10^5
\]

such numbers.

The desired fraction is

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

Choice B is \(\frac1{80}\). The answer is not \(\frac1{100}\), because the first digit is not equally likely to be any of 10 digits; it has only 8 valid possibilities.

ANSWER 4: B

Problem 5:

Let Amy, Jan, and Toy initially have

\[
a,\quad j,\quad 36
\]

dollars, respectively.

First, Amy doubles Jan’s money and Toy’s money.

- Amy gives Jan \(j\) dollars.
- Amy gives Toy \(36\) dollars.

After Amy’s turn:

\[
\text{Amy}=a-j-36,\qquad \text{Jan}=2j,\qquad \text{Toy}=72.
\]

Next, Jan doubles Amy’s and Toy’s amounts.

Jan gives Amy \(a-j-36\) dollars and gives Toy \(72\) dollars. Then:

\[
\text{Amy}=2(a-j-36),
\]

\[
\text{Jan}=2j-(a-j-36)-72=3j-a-36,
\]

\[
\text{Toy}=144.
\]

Finally, Toy doubles both Amy’s and Jan’s amounts. Toy must give them their current amounts, so Toy ends with

\[
144-2(a-j-36)-(3j-a-36).
\]

Simplify:

\[
144-2a+2j+72-3j+a+36
=252-a-j.
\]

We are told Toy finishes with \$36, so

\[
252-a-j=36.
\]

Therefore,

\[
a+j=216.
\]

The original total amount of money was

\[
a+j+36=216+36=252.
\]

Choice D is \(\$252\).

ANSWER 5: D

Problem 6:

We need the least possible value of \(M\), the median of the five group medians.

For \(M\) to be at most 8, at least three of the five group medians would have to be at most 8.

But if a group’s median is at most 8, then that group contains at least 3 numbers that are at most 8, since the median is the third number when the group is arranged in order.

So three such groups would require at least

\[
3\cdot 3=9
\]

different numbers that are at most 8.

However, there are only 8 numbers at most 8:

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

Therefore,

\[
M\ge 9.
\]

Now we show that \(M=9\) is possible. Make three groups such as

\[
\{1,2,3,16,17\},\qquad \{4,5,6,18,19\},\qquad \{7,8,9,20,21\}.
\]

Their medians are

\[
3,\ 6,\ 9.
\]

Put the remaining numbers into two groups of 5. Their medians will be greater than 9, so the five medians have third median equal to 9.

Thus the least possible value is

\[
9.
\]

Choice A is 9.

ANSWER 6: A

Problem 7:

Each circle has radius \(AB\), because each circle passes through the other circle’s center.

Thus triangle \(AEB\) has all three sides equal:

\[
AE=BE=AB.
\]

So triangle \(AEB\) is equilateral, and

\[
\angle AEB=60^\circ.
\]

Points \(C,A,B,D\) lie on the same line. Also,

\[
AC=AE
\]

because both are radii of the circle centered at \(A\). Thus triangle \(ACE\) is isosceles.

Since \(C,A,B\) are collinear,

\[
\angle CAE=180^\circ-60^\circ=120^\circ.
\]

Therefore the two base angles of triangle \(ACE\) are

\[
\frac{180^\circ-120^\circ}{2}=30^\circ.
\]

Similarly, triangle \(BDE\) is isosceles, and

\[
\angle BED=30^\circ.
\]

Therefore,

\[
\angle CED
=
\angle CEA+\angle AEB+\angle BED
=
30^\circ+60^\circ+30^\circ
=
120^\circ.
\]

Choice C is \(120^\circ\).

ANSWER 7: C

Problem 8:

The statements are:

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