🔬 Model Lab

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre

Problem 1:
We need integer-sided rectangles with perimeter 50 that maximize and minimize area, then find the difference.

Let sides be $l$ and $w$, positive integers. Perimeter is
$$2(l+w)=50$$
so
$$l+w=25,\quad w=25-l.$$
Area is
$$A = l w = l(25-l).$$

This is largest when $l$ and $w$ are as close as possible, smallest when as far apart as possible.

Since $l+w=25$ is odd, we cannot have $l=w$. The closest integers are 12 and 13:
$$12+13=25,\quad A_{\max}=12\times 13=156.$$
Any other pair is more unbalanced and gives smaller area, e.g. $11\times14=154$.

The most unbalanced positive integers with sum 25 are 1 and 24:
$$1+24=25,\quad A_{\min}=1\times 24=24.$$
Any other pair gives larger area.

Difference:
$$156-24=132.$$

This rules out 76, 120, 128, 136. The tempting $12\times13$ vs $1\times24$ must be computed carefully.

ANSWER 1: D

Problem 2:
We need the 9th and 10th game scores given averages are integers and each score is $<10$.

Sum of first 8 games:
$$7+4+3+6+8+3+1+5 = 37.$$

Let 9th game score be $x$, $0\le x\le 9$, integer. Average of 9 games:
$$\frac{37+x}{9}$$
must be an integer. So $37+x$ must be a multiple of 9.

$37+x$ ranges from 37 to 46. The multiples of 9 near are 36 and 45. Only 45 is in range. So
$$37+x=45,\quad x=8.$$
This is $<10$, valid. Total after 9 games is 45.

Let 10th game score be $y$, $0\le y\le 9$. Average of 10 games:
$$\frac{45+y}{10}$$
must be an integer. So $45+y$ must be a multiple of 10.

$45+y$ ranges from 45 to 54. The only multiple of 10 in range is 50. So
$$45+y=50,\quad y=5.$$

Product:
$$x\cdot y = 8\times 5 = 40.$$

Other products like 35, 48, 56, 72 would require different $x,y$ that do not make the totals divisible.

ANSWER 2: B

Problem 3:
Ratio is 2 boys for every 3 girls.

Restated: out of every $2+3=5$ students, 2 are boys.

Fraction who are boys:
$$\frac{2}{2+3}=\frac25=0.40=40\%.$$

With 30 students, that is $12$ boys and $18$ girls, checking $12:18=2:3$.

So not 60% (that's girls), not 66⅔%, not 12% or 20%.

ANSWER 3: C

Problem 4:
We need the fewest coins of 1,5,10,25 that can make any value 1 to 99 cents by exact payment.

To make 4 cents we must have pennies, since 5 is too big. To make 4 cents exactly we need at least 4 pennies. So at least 4 pennies. Take exactly 4.

With 4 pennies we can make 0-4. Let $M$ be max consecutive value we can make. Initially $M=4$. To keep coverage with no gaps, the next coin $c$ must satisfy $c\le M+1$, and then new $M$ becomes $M+c$.

After 4 pennies, $M=4$, so next coin $\le 5$. The only choice $\le5$ besides penny is a 5-cent nickel. So we must add a nickel. Then $M=4+5=9$.

Now next coin $\le10$. The largest allowed is a dime (10). Choosing the largest keeps coin count minimal. Add a dime: $M=9+10=19$.

Next coin $\le20$. Quarter 25 is $>20$ and would leave a gap, so we cannot yet use a quarter. Largest allowed is again a dime. Add a second dime: $M=19+10=29$.

Now next coin $\le30$. Quarter 25 qualifies. Add quarters to jump fastest. With $M=29$:
- 1 quarter gives $M=54$, still $<99$.
- 2 quarters give $M=79$, still $<99$.
- 3 quarters give $M=104\ge99$.

So we need 3 quarters.

Total: 4 pennies + 1 nickel + 2 dimes + 3 quarters = 10 coins. This can make any value to 99, and any fewer fails by the gap argument above.

ANSWER 4: B

Problem 5:
Start with 100% of marbles.

Give 20% to Pedro: left
$$100\%-20\%=80\%.$$

Give 10% of what is left to Ebony:
$$10\%\times 80\% = 8\% \text{ given},$$
left
$$80\%-8\%=72\%,$$
which is $0.8\times0.9=0.72$.

Give 25% of what is now left to Jimmy:
$$25\%\times72\%=18\% \text{ given},$$
left
$$72\%-18\%=54\%,$$
which is $0.8\times0.9\times0.75=0.54$.

So Gilda keeps 54% of original.

ANSWER 5: E

Problem 6:
We need distinct $a<b<c$ from 1-9 with $a+b+c=15$ and one of them is 5.

Let the set contain 5 plus $x$ and $y$ distinct from each other and from 5:
$$5+x+y=15,\quad x+y=10.$$

With $x\ne y$, $x\ne5$, $y\ne5$, from 1-9. Let $x<y$. Then $x<5$ (since $x+y=10$).

$x=1\Rightarrow y=9$: {1,5,9} valid.
$x=2\Rightarrow y=8$: {2,5,8} valid.
$x=3\Rightarrow y=7$: {3,5,7} valid.
$x=4\Rightarrow y=6$: {4,5,6} valid.

No other $x$ works. So there are 4 such sets.

ANSWER 6: B

Problem 7:
Two 99-digit numbers:
$$N_1 = 3030\ldots303,\quad N_2 = 5050\ldots505$$
alternating 3,0 and 5,0, beginning and ending with 3 and 5 respectively.

Units digit $B$: $3\times5=15$, so $B=5$.

Thousands digit $A$: only the last 4 digits of each factor affect the last 4 digits of the product, because modulo 10000,
$$(H\cdot10000+L)(H'\cdot10000+L')\equiv L L' \pmod{10000}.$$

Find last 4 digits. $N_1$ has odd positions 3, even 0. Positions 96,97,98,99 from the left are 0,3,0,3. So last four digits are 0303 = 303. Similarly $N_2$ ends in 0505 = 505.

Multiply:
$$303\times505 = 303\times500+303\times5 = 151500+1515 = 153015.$$

Mod 10000 this is 3015. So the product ends ...3015. Thus units=5, tens=1, hundreds=0, thousands=3. So $A=3$, $B=5$.

Sum:
$$A+B=3+5=8.$$

ANSWER 7: D

Problem 8:
Find $N$ with
$$N\equiv4\pmod6,\quad N\equiv3\pmod5,$$
smallest such, then find $N\bmod7$.

Write $N=6k+4$. Need:
$$6k+4\equiv3\pmod5.$$
$$6k\equiv-1\equiv4\pmod5.$$
Since $6\equiv1\pmod5$, $k\equiv4\pmod5$. Smallest non-negative $k$ is 4.

Then
$$N=6\cdot4+4=28.$$
Check: $28=4\times6+4$, $28=5\times5+3$. Good.

Now divide by 7:
$$28\equiv0\pmod7.$$
Remainder 0.

ANSWER 8: A

Problem 9:
Speech length between $1/2$ hour and $3/4$ hour at 150 words per minute.

Convert:
$$\frac12\text{ hr}=30\text{ min},\quad \frac34\text{ hr}=45\text{ min}.$$

Word counts:
$$\text{min}=30\times150=4500,\quad \text{max}=45\times150=6750.$$

We need a count between 4500 and 6750 inclusive.

Choices: 2250 too small, 3000 too small, 4200 too small, 4350 too small (below 4500), 5650 is between 4500 and 6750. So only 5650 is appropriate.

ANSWER 9: E

Problem 10:
Three positive integers equally spaced, middle 15, largest = 4 $\times$ smallest.

Let smallest $=s$, largest $=L$. Equally spaced means middle is the average:
$$15-s = L-1

Problem 1:
We are asked to find the difference between the largest and smallest possible areas of a rectangle with integer side lengths and a perimeter of 50 units.

1. Let the side lengths of the rectangle be integers $a$ and $b$, with $a \le b$.
2. The perimeter is given by $2(a + b) = 50$, which simplifies to:
   $$a + b = 25$$
3. Since $a$ and $b$ are positive integers, the possible values for $a$ range from $1$ to $12$:
   - The smallest area occurs when the side lengths are as far apart as possible:
     $$a = 1, \quad b = 24 \implies \text{Area}_{\min} = 1 \times 24 = 24$$
   - The largest area occurs when the side lengths are as close to each other as possible:
     $$a = 12, \quad b = 13 \implies \text{Area}_{\max} = 12 \times 13 = 156$$
4. The difference between the largest and smallest areas is:
   $$156 - 24 = 132$$

ANSWER 1: D

---

Problem 2:
Theresa's scores in her first 8 games are $7, 4, 3, 6, 8, 3, 1,$ and $5$. We want to find the product of her scores in the ninth and tenth games, knowing each score is fewer than 10 and each resulting points-per-game average is an integer.

1. First, calculate the total points scored in the first 8 games:
   $$\text{Sum}_8 = 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37$$
2. Let $x$ be her score in the ninth game, with $0 \le x < 10$. The total for 9 games is $37 + x$, which must be divisible by 9:
   - The multiples of 9 are $9, 18, 27, 36, 45, 54, \dots$
   - The only multiple of 9 reachable with $0 \le x < 10$ is $45$:
     $$37 + x = 45 \implies x = 8$$
3. Let $y$ be her score in the tenth game, with $0 \le y < 10$. The total for 10 games is $45 + y$, which must be divisible by 10:
   - The multiples of 10 are $10, 20, 30, 40, 50, 60, \dots$
   - The only multiple of 10 reachable with $0 \le y < 10$ is $50$:
     $$45 + y = 50 \implies y = 5$$
4. The product of her scores in the ninth and tenth games is:
   $$x \times y = 8 \times 5 = 40$$

ANSWER 2: B

---

Problem 3:
We are asked to find the percentage of students in Ms. Johnson's class who are boys, given that there are 2 boys for every 3 girls.

1. The ratio of boys to girls is $2 : 3$.
2. This means that out of every $2 + 3 = 5$ students, 2 are boys.
3. The fraction of students who are boys is:
   $$\frac{2}{2 + 3} = \frac{2}{5}$$
4. Converting this fraction to a percentage:
   $$\frac{2}{5} \times 100\% = 40\%$$
   (Note: The total of 30 students confirms this, with $\frac{2}{5} \times 30 = 12$ boys, and $\frac{12}{30} = 40\%$.)

ANSWER 3: C

---

Problem 4:
We want to find the minimum number of coins (using pennies = 1¢, nickels = 5¢, dimes = 10¢, quarters = 25¢) needed to make every amount from 1¢ to 99¢.

1. To form amounts ending in 1, 2, 3, and 4 cents, we must be able to add 1, 2, 3, or 4 cents to any multiple of 5. This requires at least 4 pennies.
2. The remaining coins must be able to form all multiples of 5 from 0¢ to 95¢:
   - If we use 3 quarters (75¢), we need to be able to form 0¢, 5¢, 10¢, 15¢, and 20¢ using the remaining coins.
   - Using 1 nickel and 2 dimes allows us to make:
     - $0¢$
     - $5¢ = 1\text{ nickel}$
     - $10¢ = 1\text{ dime}$
     - $15¢ = 1\text{ dime} + 1\text{ nickel}$
     - $20¢ = 2\text{ dimes}$
   - Combining $\{0, 5, 10, 15, 20\}¢$ with multiples of 25¢ up to 75¢ ($0, 25, 50, 75$) can generate every multiple of 5 up to $75 + 20 = 95¢$.
   - This uses $3\text{ quarters} + 2\text{ dimes} + 1\text{ nickel} = 6$ coins.
3. Can we use fewer than 6 non-penny coins?
   - With 2 or fewer quarters, the maximum amount reachable with 5 non-penny coins would not cover all necessary combinations up to 95¢.
   - With 3 quarters and only 2 coins left, two coins can form at most $2^2 - 1 = 3$ non-zero values, which cannot cover the 4 values $\{5, 10, 15, 20\}¢$.
   - Therefore, at least 6 non-penny coins are required.
4. Total minimum number of coins:
   $$4\text{ pennies} + 1\text{ nickel} + 2\text{ dimes} + 3\text{ quarters} = 10\text{ coins}$$

ANSWER 4: B

---

Problem 5:
We want to find the percentage of marbles Gilda has left after giving away portions of her marbles in three stages.

1. Start with $100\%$ of the marbles.
2. After giving $20\%$ to Pedro, she retains:
   $$100\% \times (1 - 0.20) = 80\%$$
3. After giving $10\%$ of the remainder to Ebony, she retains:
   $$80\% \times (1 - 0.10) = 80\% \times 0.90 = 72\%$$
4. After giving $25\%$ of the new remainder to Jimmy, she retains:
   $$72\% \times (1 - 0.25) = 72\% \times 0.75 = 54\%$$

ANSWER 5: E

---

Problem 6:
We want to find the number of three-element subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that sum to 15 and contain the number 5.

1. Let the three numbers in the set be $\{a, b, 5\}$, where $a$ and $b$ are distinct elements from $\{1, 2, 3, 4, 6, 7, 8, 9\}$.
2. The sum of the elements must be 15:
   $$a + b + 5 = 15 \implies a + b = 10$$
3. Without loss of generality, assume $a < b$. We list the pairs of distinct integers from $\{1, 2, 3, 4, 6, 7, 8, 9\}$ that sum to 10:
   - $1 + 9 = 10 \implies \{1, 9, 5\}$
   - $2 + 8 = 10 \implies \{2, 8, 5\}$
   - $3 + 7 = 10 \implies \{3, 7, 5\}$
   - $4 + 6 = 10 \implies \{4, 6, 5\}$
   (Note that $5 + 5 = 10$ is not allowed because all three numbers in the set must be distinct.)
4. There are 4 such sets.

ANSWER 6: B

---

Problem 7:
We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product of two 99-digit numbers:
$$X = 303,030,303,\dots,030,303 \quad \text{and} \quad Y = 505,050,505,\dots,050,505$$

1. The thousands digit and units digit of the product depend entirely on the last 4 digits of the product (i.e., the product modulo $10,000$).
2. Looking at the last 4 digits of each number:
   - $X \equiv 0303 = 303 \pmod{10000}$
   - $Y \equiv 0505 = 505 \pmod{10000}$
3. Multiply these two 4-digit values:
   $$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
4. Taking this product modulo $10,000$:
   $$153015 \equiv 3015 \pmod{10000}$$
5. The last four digits are $3, 0, 1, 5$:
   - The thousands di

# Problem 1:
A rectangle with perimeter 50 and integer sides l and w satisfies:
- l + w = 25 (since 2(l + w) = 50)

**Smallest area:** Make the rectangle as "skinny" as possible: 1 × 24, giving area = 24.

**Largest area:** For a fixed perimeter, area is maximized when sides are as equal as possible. With l + w = 25, the closest integer pair is 12 × 13, giving area = 156.

**Difference:** 156 − 24 = 132.

(Checking other options: 156 − 24 is not 120, 128, 136, or 76 — the tempting error is picking 13 × 12 = 156 and subtracting 2 × 23 = 46 to get 110, which isn't even listed; the correct minimum is 24.)

ANSWER 1: D

# Problem 2:
Total points in the first 8 games: 7 + 4 + 3 + 6 + 8 + 3 + 1 + 5 = 37.

**Ninth game:** Let x < 10 be her score. We need (37 + x) divisible by 9.
- 37 + x ≡ 0 (mod 9) → x ≡ −37 ≡ −37 + 45 = 8 (mod 9)
- So x = 8. New total: 37 + 8 = 45, and 45 ÷ 9 = 5. ✓

**Tenth game:** Let y < 10 be her score. We need (45 + y) divisible by 10.
- 45 + y ≡ 0 (mod 10) → y = 5. New total: 50, and 50 ÷ 10 = 5. ✓

**Product:** 8 × 5 = 40.

ANSWER 2: B

# Problem 3:
The ratio of boys to girls is 2 : 3, so boys make up 2/(2+3) = 2/5 of the class.

Fraction of boys: 2/5 = 40%.

(As a check with 30 students: 2/5 × 30 = 12 boys and 18 girls — ratio 12:18 = 2:3. ✓)

ANSWER 3: C

# Problem 4:
To pay **any** amount from 1¢ to 99¢, Freddie needs coins that can build every value in that range.

- **Pennies:** Amounts 1–4¢ require pennies (no other coin is that small), so he needs **4 pennies**.
- **Nickel:** 5¢ cannot be made without a nickel, so he needs **1 nickel** (nickel + pennies covers 6–9¢).
- **Dimes:** With 1 nickel and 4 pennies he covers 0–9¢. Adding dimes extends coverage by 10 each. **2 dimes** cover up to 29¢ (all values 0–29 reachable).
- **Quarters:** Each quarter extends the contiguous range by 25: 0–54, 0–79, 0–104. **3 quarters** are needed to reach past 99¢ (only 2 quarters reaches 79¢).

**Total:** 4 + 1 + 2 + 3 = **10 coins**: {25, 25, 25, 10, 10, 5, 1, 1, 1, 1}.

Check a few tricky amounts: 40¢ = 25 + 10 + 5 ✓; 85¢ = 25 + 25 + 25 + 10 ✓; 99¢ = all coins ✓; 30¢ = 25 + 5 ✓.

ANSWER 4: B

# Problem 5:
Start with 100% of the marbles and track what remains after each gift:

- After giving Pedro 20%: Gilda keeps 100% − 20% = 80%.
- After giving Ebony 10% of what's left: Gilda keeps 90% of 80% = 0.90 × 0.80 = 72%.
- After giving Jimmy 25% of what's left: Gilda keeps 75% of 72% = 0.75 × 0.72 = 0.54 = **54%**.

(Check with 100 marbles: gives 20 → 80; gives 8 → 72; gives 18 → 54 left. ✓)

ANSWER 5: E

# Problem 6:
We need 3-element sets of *different* numbers from {1, ..., 9} with sum 15 that contain a 5.

If 5 is in the set, the other two numbers must be distinct, from {1, 2, 3, 4, 6, 7, 8, 9}, and sum to 15 − 5 = 10.

List all pairs of distinct numbers summing to 10:
- {1, 9}
- {2, 8}
- {3, 7}
- {4, 6}
- ({5, 5} is invalid — numbers must be different)

That gives 4 sets: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.

ANSWER 6: B

# Problem 7:
Write each number as a multiple of a "repeating ones" pattern:
- 303,030,...,303 (99 digits) = 3 · R, where R = 101,010,...,101 (the digit 1 in every other position).
- 505,050,...,505 (99 digits) = 5 · R.

So the product = 15 · R², where R = 10⁹⁸ + 10⁹⁶ + ... + 10² + 1.

**Units digit B:** The units digit of the product comes from the units digits: 3 × 5 = 15, so **B = 5**.

**Thousands digit A:** Work modulo 10000. In R, every term 10^{2k} with k ≥ 2 is a multiple of 10⁴ = 10000, so it vanishes mod 10000:
- R ≡ 10² + 1 = 101 (mod 10000).
- R² ≡ 101² = 10201 ≡ 201 (mod 10000).
- Product ≡ 15 × 201 = 3015 (mod 10000).

So the last four digits are 3015, and the **thousands digit A = 3**.

**Sum:** A + B = 3 + 5 = 8.

ANSWER 7: D

# Problem 8:
Let N be the number of coins. The conditions are:
- N ≡ 4 (mod 6)
- N ≡ 3 (mod 5)

Numbers that leave remainder 4 when divided by 6: 4, 10, 16, 22, 28, 34, ...
Check remainder mod 5:
- 4 mod 5 = 4 ✗
- 10 mod 5 = 0 ✗
- 16 mod 5 = 1 ✗
- 22 mod 5 = 2 ✗
- 28 mod 5 = 3 ✓

So the smallest N is 28. Check: 28 = 6(4) + 4 ✓ and 28 = 5(5) + 3 ✓.

**Divide by 7:** 28 ÷ 7 = 4 with remainder **0**.

ANSWER 8: A

# Problem 9:
The speech must last between 30 minutes (one-half hour) and 45 minutes (three-quarters of an hour).

At 150 words per minute:
- Minimum words: 30 × 150 = 4500
- Maximum words: 45 × 150 = 6750

So the word count must be between 4500 and 6750.

Check each choice:
- 2250 → only 15 minutes ✗
- 3000 → only 20 minutes ✗
- 4200 → only 28 minutes ✗
- 4350 → only 29 minutes ✗
- 5650 → 5650 ÷ 150 ≈ 37.7 minutes, which is between 30 and 45 ✓

ANSWER 9: E

# Problem 10:
Let the three equally spaced numbers be a, 15, and 4a, where a is the smallest.

Since they are equally spaced, the gap between consecutive numbers is the same:
- 15 − a = 4a − 15

Solve:
- 15 + 15 = 4a + a
- 30 = 5a
- a = 6

**Check:** The numbers are 6, 15, 24 — equally spaced (gaps of 9), and 24 = 4 × 6. ✓

ANSWER 10: C

# Problem 1:
Perimeter 50 means 2(l+w) = 50, so l + w = 25 with l, w positive integers.

- Area is largest when the sides are as close as possible: 12 × 13 = 156.
- Area is smallest when the sides are as far apart as possible: 1 × 24 = 24.

Difference: 156 − 24 = **132**.

(Tempting wrong answers come from using 12×12=144 or 25 as a side, but sides must sum to 25 and be positive.)

ANSWER 1: D

# Problem 2:
Sum of first 8 games: 7+4+3+6+8+3+1+5 = 37.

Ninth game: 37 + x must be divisible by 9, with x < 10. Since 37 ≡ 1 (mod 9), we need x ≡ 8 (mod 9), so x = 8. Total = 45 (average 5).

Tenth game: 45 + y must be divisible by 10, with y < 10. So y = 5. Total = 50 (average 5).

Product = 8 × 5 = **40**.

ANSWER 2: B

# Problem 3:
Ratio boys : girls = 2 : 3, so 5 parts total. With 30 students, each part is 6.
Boys = 2 × 6 = 12, so percent boys = 12/30 = 40%.

(60% is the girls' share; 66⅔% comes from mistakenly using 2/3.)

ANSWER 3: C

# Problem 4:
He must be able to make every amount from 1¢ to 99¢.

Try: 4 pennies, 1 nickel, 2 dimes, 3 quarters = 4+1+2+3 = 10 coins.
- Pennies handle the 1–4 remainder.
- Nickel + pennies handle 5–9.
- Dimes + above handle 10–24.
- Quarters + above handle up to 75+20+5+4 = 99. ✓

Fewer than 10 is impossible: with only 9 coins you can't cover all residues (you need at least 4 pennies, at least 1 nickel, at least 2 dimes, and at least 3 quarters to reach 99).

Answer: **10**.

ANSWER 4: B

# Problem 5:
Start with 100%.
- After Pedro: 80% remains (multiply by 0.8).
- After Ebony: 0.8 × 0.9 = 0.72 remains.
- After Jimmy: 0.72 × 0.75 = 0.54.

So Gilda keeps 54%. (Subtracting 20+10+25 = 55% to get 45% is the trap — the percentages apply to shrinking amounts.)

ANSWER 5: E

# Problem 6:
We need three distinct numbers from 1–9 summing to 15 that include 5. The other two must be distinct, ≠5, and sum to 10:
(1,9), (2,8), (3,7), (4,6).

That's 4 sets. (5,5 is not allowed since numbers must differ.)

ANSWER 6: B

# Problem 7:
The 99-digit number 3030…303 equals 3S, where S = 1 + 100 + 100² + … + 100⁴⁹.
Likewise 5050…505 = 5S.

Product = 15S².

Now S² = Σ cₙ·100ⁿ with c₀ = 1, c₁ = 2, and all other terms multiples of 100² = 10⁴.
So S² ≡ 1 + 200 = 201 (mod 10 000).

Then 15S² ≡ 15 × 201 = 3015 (mod 10 000).

Last four digits are 3015: thousands digit A = 3, units digit B = 5.
A + B = **8**.

ANSWER 7: D

# Problem 8:
Let n be the number of coins: n ≡ 4 (mod 6) and n ≡ 3 (mod 5).

Numbers ≡ 4 mod 6: 4, 10, 16, 22, 28, …
Their remainders mod 5: 4, 0, 1, 2, 3 → first match is n = 28.

28 ÷ 7 = 4 exactly, so the remainder is **0**.

ANSWER 8: A

# Problem 9:
Time must be between 30 and 45 minutes at 150 words/minute:
- Minimum words: 150 × 30 = 4500
- Maximum words: 150 × 45 = 6750

Which choice lies in [4500, 6750]? 2250, 3000, 4200, 4350 are all too small; 5650 works.

ANSWER 9: E

# Problem 10:
Equally spaced numbers: 15 − d, 15, 15 + d.
Given 15 + d = 4(15 − d):

15 + d = 60 − 4d → 5d = 45 → d = 9.

So the numbers are 6, 15, 24, and indeed 24 = 4 × 6. Smallest = **6**.

ANSWER 10: C

Problem 1:

We need rectangles with integer side lengths and perimeter \(50\).

If the side lengths are \(l\) and \(w\), then

\[
2l+2w=50
\]

so

\[
l+w=25.
\]

The area is \(lw\).

- The smallest possible positive integer side length is \(1\), giving sides \(1\) and \(24\):

\[
1\cdot 24=24.
\]

- For a fixed sum, the area is largest when the two numbers are as close together as possible. The closest integers adding to \(25\) are \(12\) and \(13\):

\[
12\cdot 13=156.
\]

Thus the difference is

\[
156-24=132.
\]

Choice D is correct. Choices such as \(128\) or \(136\) could result from using incorrect side pairs, but \(1,24\) and \(12,13\) give the true minimum and maximum areas.

ANSWER 1: D

Problem 2:

First find Theresa’s total after eight games:

\[
7+4+3+6+8+3+1+5=37.
\]

Let her ninth-game score be \(x\). Her nine-game average is an integer, so

\[
37+x
\]

must be divisible by \(9\).

Since

\[
37\equiv 1\pmod 9,
\]

we need

\[
x\equiv 8\pmod 9.
\]

Because she scored fewer than \(10\) points, the only possibility is

\[
x=8.
\]

Her total after nine games is

\[
37+8=45.
\]

Let her tenth-game score be \(y\). Her ten-game average is an integer, so

\[
45+y
\]

must be divisible by \(10\). Since \(y<10\), we need

\[
y=5.
\]

The requested product is

\[
8\cdot 5=40.
\]

Choice B is correct. For example, \(35\) would come from scores \(7\) and \(5\), but \(7\) does not make the first nine-game total divisible by \(9\).

ANSWER 2: B

Problem 3:

The ratio of boys to girls is

\[
2:3.
\]

That means there are

\[
2+3=5
\]

equal ratio parts total, and boys make up \(2\) of those \(5\) parts.

Thus the fraction who are boys is

\[
\frac25.
\]

Converting to a percent:

\[
\frac25=0.4=40\%.
\]

So \(40\%\) of the class are boys.

Choice C is correct. The \(60\%\) choice is the percentage of girls, not boys.

ANSWER 3: C

Problem 4:

Freddie needs a collection of pennies, nickels, dimes, and quarters that can make every amount from \(1\) cent through \(99\) cents.

To make \(1\), \(2\), \(3\), and \(4\) cents, he must have four pennies.

So far he has:

\[
1+1+1+1=4\text{ cents}.
\]

Then he can add a nickel, giving coverage through

\[
4+5=9\text{ cents}.
\]

Next, add a dime. Since he can already make every amount through \(9\), adding a dime lets him make every amount through

\[
9+10=19.
\]

Add another dime:

\[
19+10=29.
\]

Now quarters can be used, because he can make all amounts through \(29\), including the amounts needed to combine with a quarter. Add three quarters:

\[
29+25+25+25=104.
\]

Thus the coins are:

- 4 pennies,
- 1 nickel,
- 2 dimes,
- 3 quarters.

That is

\[
4+1+2+3=10
\]

coins, and they can make every amount from \(1\) through \(99\) cents.

Why can’t 9 coins work? Building the greatest possible usable value while avoiding gaps gives:

\[
1,1,1,1,5,10,10,25,25,
\]

whose total is only

\[
1+1+1+1+5+10+10+25+25=79.
\]

So 9 coins cannot even have enough usable coverage to reach \(99\) cents.

Choice B is correct.

ANSWER 4: B

Problem 5:

Suppose Gilda starts with \(100\%\) of her marbles.

After giving Pedro \(20\%\), she has

\[
100\%-20\%=80\%.
\]

Then she gives Ebony \(10\%\) of what remains, so she keeps \(90\%\) of the \(80\%\):

\[
0.90\cdot 80\%=72\%.
\]

Finally, she gives Jimmy \(25\%\) of what remains, so she keeps \(75\%\) of the \(72\%\):

\[
0.75\cdot72\%=54\%.
\]

Therefore, Gilda has \(54\%\) of her original marbles left.

Choice E is correct. A common mistake is to subtract \(20\%+10\%+25\%=55\%\), but the later percentages are taken from smaller remaining amounts.

ANSWER 5: E

Problem 6:

The set must contain \(5\), and the total of the three distinct numbers must be \(15\).

So the other two numbers must add to

\[
15-5=10.
\]

Using different numbers from \(\{1,2,3,4,5,6,7,8,9\}\), the pairs that total \(10\) are:

\[
1+9,
\]

\[
2+8,
\]

\[
3+7,
\]

\[
4+6.
\]

The pair \(5+5\) is not allowed because the numbers must be different.

Thus the sets are:

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

There are \(4\) such sets.

Choice B is correct.

ANSWER 6: B

Problem 7:

We need the thousands digit and units digit of the product.

Only the last four digits of each factor matter for finding the last four digits of the product.

The first number ends in

\[
\ldots 030,303,
\]

so its last four digits are

\[
0303=303.
\]

The second number ends in

\[
\ldots 050,505,
\]

so its last four digits are

\[
0505=505.
\]

Now multiply:

\[
303\cdot 505
\]

\[
=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\). Therefore:

- the thousands digit is \(3\),
- the units digit is \(5\).

Their sum is

\[
3+5=8.
\]

Choice D is correct.

ANSWER 7: D

Problem 8:

Let \(N\) be the number of coins.

When divided among six people, four are left over:

\[
N\equiv 4\pmod 6.
\]

When divided among five people, three are left over:

\[
N\equiv 3\pmod 5.
\]

Numbers that leave remainder \(4\) when divided by \(6\) are:

\[
4,10,16,22,28,34,\ldots
\]

Among these, the first one that leaves remainder \(3\) when divided by \(5\) is

\[
28,
\]

because

\[
28\div 5
\]

leaves remainder \(3\).

Now divide \(28\) by \(7\):

\[
28\div 7=4
\]

with remainder \(0\).

So no coins are left over when the coins are equally divided among seven people.

Choice A is correct.

ANSWER 8: A

Problem 9:

Julie speaks at \(150\) words per minute.

Her speech must last from one-half hour to three-quarters of an hour:

\[
30\text{ minutes to }45\text{ minutes}.
\]

At \(150\) words per minute, the smallest appropriate word count is

\[
30\cdot 150=4500.
\]

The largest appropriate word count is

\[
45\cdot 150=6750.
\]

So the speech must contain between \(4500\) and \(6750\) words.

Check the choices:

- \(2250\) words is too short.
- \(3000\) words is too short.
- \(4200\) words is too short.
- \(4350\) words is too short.
- \(5650\) words is between \(4500\) and \(6750\).

Thus \(5650\) is appropriate.

Choice E is corre