🔬 Model Lab

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

---

### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

---

### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

---

### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

---

### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
---
### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
---
### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
---
### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
---
### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
---
### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
---
### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
---
### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
---
### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
---
### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
---
### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
---
### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A

I'll solve each problem step-by-step.

**Problem 1:**

Given: a ∗ b means 3a − b
Find: x if 2 ∗ (5 ∗ x) = 1

First, find 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now find 2 ∗ (5 ∗ x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

**Problem 2:**

Maria works 8 hours (not including 45-minute lunch).
- Starts: 7:25 A.M.
- Takes lunch at noon

From 7:25 A.M. to noon = 4 hours 35 minutes of work

Remaining work needed: 8 − 4 hours 35 minutes = 3 hours 25 minutes

After lunch at noon, add 3 hours 25 minutes:
12:00 P.M. + 3 hours 25 minutes = 3:25 P.M.

Plus the 45-minute lunch break:
3:25 P.M. + 45 minutes = 4:10 P.M.

ANSWER 2: C

**Problem 3:**

Find GCD and LCM of 180 and 594.

Prime factorization:
- 180 = 2² × 3² × 5
- 594 = 2 × 3³ × 11

GCD(180, 594) = 2 × 3² = 18

LCM(180, 594) = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Ratio: 5940/18 = 330

ANSWER 3: C

**Problem 4:**

Starting with $2 on Monday, track all possible amounts for each day (gain $3 or double):

Monday: $2

Tuesday: 2+3=$5 or 2×2=$4

Wednesday: 
- From $5: 5+3=$8 or 5×2=$10
- From $4: 4+3=$7 or 4×2=$8
- Distinct amounts: $7, $8, $10

Thursday:
- From $7: 7+3=$10 or 7×2=$14
- From $8: 8+3=$11 or 8×2=$16
- From $10: 10+3=$13 or 10×2=$20
- Distinct amounts: $10, $11, $13, $14, $16, $20

Six different amounts possible.

ANSWER 4: D

**Problem 5:**

Three equally spaced integers: a, a+d, a+2d

Given:
- a + (a+d) = 40 → 2a + d = 40
- (a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtract first from second:
2d = 20 → d = 10

From first equation:
2a + 10 = 40 → a = 15

The three numbers: 15, 25, 35
Sum: 15 + 25 + 35 = 75

ANSWER 5: B

**Problem 6:**

Total coins: 2020, with at least 1 penny and 1 nickel.

Maximum value: 2019 nickels + 1 penny = 2019(5) + 1 = 10,095 + 1 = 10,096 cents

Minimum value: 1 nickel + 2019 pennies = 5 + 2019 = 2,024 cents

Difference: 10,096 − 2,024 = 8,072 cents

ANSWER 6: C

**Problem 7:**

Isosceles triangle has at least two equal angles.

Case 1: Two angles are 70° each
Third angle: 180 − 70 − 70 = 40°, so x = 40

Case 2: One angle is 70°, the other two are equal to x
70 + x + x = 180 → x = 55

Case 3: Two angles equal x, one is 70°
x + x + 70 = 180 → x = 55 (duplicate)

Wait, let me reconsider: the problem says "two angles measure 70° and x°", meaning we're told about these two angles.

- If both 70° and x° are the equal pair: x = 70, third = 40
- If 70° appears twice: x = 180 − 140 = 40
- If x appears twice with a 70°: 2x + 70 = 180, so x = 55

Possible values: 40, 55, 70
Sum: 40 + 55 + 70 = 165

ANSWER 7: D

**Problem 8:**

2 ÷ (1 − 2/3) = 2 ÷ (1/3) = 2 × 3 = 6

ANSWER 8: E

**Problem 9:**

Number leaves remainder 2 when divided by 3, 4, 5, 6.
This means: n ≡ 2 (mod 3, 4, 5, 6)
So: n − 2 is divisible by 3, 4, 5, 6

LCM(3,4,5,6) = 60
n − 2 = 60k, so n = 60k + 2

For k = 1: n = 62 (between 60 and 79)

ANSWER 9: B

**Problem 10:**

Find 3-digit numbers with digit product = 24.

Factorizations of 24 into three single digits:
1×3×8, 1×4×6, 2×2×6, 2×3×4, 1×1×24 (invalid)

- {1,3,8}: 3! = 6 arrangements
- {1,4,6}: 3! = 6 arrangements
- {2,2,6}: 3!/2! = 3 arrangements
- {2,3,4}: 3! = 6 arrangements

Total: 6+6+3+6 = 21

ANSWER 10: D

**Problem 11:**

Ben thinks of 6:
Ben's operation: (6+1)×2 = 7×2 = 14
Sue receives: 14
Sue's operation: (14−1)×2 = 13×2 = 26

ANSWER 11: C

**Problem 12:**

10001 is odd, so one prime must be 2.
10001 − 2 = 9999

Is 9999 prime? 9999 = 9×1111 = 3²×11×101 (not prime)

Check if 10001 can be written as sum of two odd primes:
Need p + q = 10001 where both are prime.

Testing: 10001 = 73 × 137 (both prime factors), so 10001 is composite.

By checking, there are no two primes summing to 10001.

ANSWER 12: A

Problem 1:

We are told that  
\[
a*b = 3a-b.
\]
We need to solve
\[
2*(5*x)=1.
\]

First compute the inside expression:
\[
5*x = 3(5)-x = 15-x.
\]

Now use that in the outer expression:
\[
2*(15-x)=3(2)-(15-x)=6-15+x=x-9.
\]

So we solve
\[
x-9=1 \implies x=10.
\]

This matches choice D.

ANSWER 1: D

---

Problem 2:

Maria works 8 hours, but lunch time is not included in those 8 hours. So we add:
- 8 hours of work
- 45 minutes of lunch

She starts working at 7:25 A.M.

From 7:25 A.M. to 12:00 noon is:
\[
4\text{ hours }35\text{ minutes}.
\]

By noon, she has worked 4 hours 35 minutes, so she still needs:
\[
8\text{ hr }-4\text{ hr }35\text{ min}=3\text{ hr }25\text{ min}.
\]

After lunch at noon, add her remaining 3 hours 25 minutes:
\[
12:00 + 3:25 = 3:25.
\]

Now include the 45-minute lunch break:
\[
3:25 + 0:45 = 4:10.
\]

So her workday ends at 4:10 P.M.

ANSWER 2: C

---

Problem 3:

We want the ratio
\[
\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}.
\]

Factor both numbers:

\[
180=2^2\cdot 3^2\cdot 5
\]
\[
594=2\cdot 3^3\cdot 11
\]

Greatest common factor:
take the smallest power of each shared prime:
\[
\gcd(180,594)=2^1\cdot 3^2=18.
\]

Least common multiple:
take the largest power of each prime:
\[
\text{lcm}(180,594)=2^2\cdot 3^3\cdot 5\cdot 11=4\cdot 27\cdot 55=5940.
\]

Now divide:
\[
\frac{5940}{18}=330.
\]

So the answer is 330.

ANSWER 3: C

---

Problem 4:

Taye starts with \$2 on Monday. For each of the next 3 days, he either:
- adds 3, or
- doubles the amount.

We list all possible amounts after 3 days.

Start: 2

After 1 day:
- \(2+3=5\)
- \(2\cdot 2=4\)

So possible amounts: \(4,5\)

After 2 days:
From 4:
- \(4+3=7\)
- \(4\cdot 2=8\)

From 5:
- \(5+3=8\)
- \(5\cdot 2=10\)

Possible amounts: \(7,8,10\)

After 3 days:
From 7:
- \(7+3=10\)
- \(7\cdot 2=14\)

From 8:
- \(8+3=11\)
- \(8\cdot 2=16\)

From 10:
- \(10+3=13\)
- \(10\cdot 2=20\)

Distinct amounts are:
\[
10,11,13,14,16,20
\]
That is 6 different amounts.

ANSWER 4: D

---

Problem 5:

Let the three equally spaced integers be
\[
a,\ a+d,\ a+2d.
\]

We are told:
\[
a+(a+d)=40
\]
and
\[
(a+d)+(a+2d)=60.
\]

Simplify:
\[
2a+d=40
\]
\[
2a+3d=60
\]

Subtract the first equation from the second:
\[
(2a+3d)-(2a+d)=60-40
\]
\[
2d=20
\]
\[
d=10.
\]

Plug into \(2a+d=40\):
\[
2a+10=40
\]
\[
2a=30
\]
\[
a=15.
\]

So the numbers are:
\[
15,\ 25,\ 35.
\]

Their sum is:
\[
15+25+35=75.
\]

ANSWER 5: B

---

Problem 6:

Ricardo has 2020 coins, each a penny or nickel, with at least one of each.

Let \(p\) be the number of pennies and \(n\) the number of nickels. Then:
\[
p+n=2020,\quad p\ge1,\quad n\ge1.
\]

Total value in cents:
\[
1p+5n.
\]
Substitute \(p=2020-n\):
\[
2020-n+5n=2020+4n.
\]

So the amount depends on \(n\).

- Least amount: use as few nickels as possible, so \(n=1\):
\[
2020+4(1)=2024.
\]

- Greatest amount: use as many nickels as possible, so \(n=2019\):
\[
2020+4(2019)=2020+8076=10096.
\]

Difference:
\[
10096-2024=8072.
\]

ANSWER 6: C

---

Problem 7:

An isosceles triangle has two equal angles. One angle is \(70^\circ\), and another is \(x^\circ\).

We consider all possible ways the \(70^\circ\) angle can fit into an isosceles triangle.

### Case 1: The two equal angles are \(70^\circ\) and \(70^\circ\)
Then the third angle is:
\[
180-70-70=40^\circ
\]
So \(x=70\) or \(x=40\) depending on which angle is labeled \(x\). But since one angle is already 70, the possible \(x\) here is \(70\) or \(40\).

### Case 2: The equal angles are \(70^\circ\) and \(x^\circ\)
Then the triangle angles are:
\[
70,\ x,\ x
\]
So
\[
70+2x=180
\]
\[
2x=110
\]
\[
x=55.
\]

Thus the possible values of \(x\) are:
\[
40,\ 55,\ 70.
\]

Their sum is:
\[
40+55+70=165.
\]

ANSWER 7: D

---

Problem 8:

Compute:
\[
\frac{2}{1-\frac23}.
\]

First simplify the denominator:
\[
1-\frac23=\frac13.
\]

So:
\[
\frac{2}{\frac13}=2\cdot 3=6.
\]

ANSWER 8: E

---

Problem 9:

A number leaves remainder 2 when divided by 3, 4, 5, and 6.

That means the number is 2 more than a multiple of all of them, so:
\[
n-2
\]
must be divisible by 3, 4, 5, and 6.

We need the least common multiple of 3, 4, 5, and 6:
\[
\text{lcm}(3,4,5,6)=60.
\]

So the smallest such number is:
\[
60+2=62.
\]

Now 62 lies between 60 and 79.

ANSWER 9: B

---

Problem 10:

We want 3-digit positive integers whose digits multiply to 24.

Let the digits be \(a,b,c\), each from 1 to 9, with
\[
abc=24.
\]

We list digit triples using factors of 24 that are single digits.

Possible factor sets:
- \(1,3,8\)
- \(1,4,6\)
- \(2,2,6\)
- \(2,3,4\)

Now count permutations:

- \(1,3,8\): all distinct \(\Rightarrow 3!=6\)
- \(1,4,6\): all distinct \(\Rightarrow 6\)
- \(2,2,6\): two equal \(\Rightarrow \frac{3!}{2!}=3\)
- \(2,3,4\): all distinct \(\Rightarrow 6\)

Total:
\[
6+6+3+6=21.
\]

ANSWER 10: D

---

Problem 11:

The teacher says:

1. Think of a number.
2. Add 1.
3. Double it.

So if Ben thinks of 6, his answer is:
\[
2(6+1)=14.
\]

Sue then:
1. subtracts 1 from the number she is given,
2. doubles the result.

So Sue gets:
\[
2(14-1)=2\cdot 13=26.
\]

ANSWER 11: C

---

Problem 12:

We want the number of ways to write 10001 as the sum of two primes.

Since 10001 is odd, one prime must be 2 and the other must be 9999 if the sum is to be odd. But 9999 is not prime, so that doesn't work.

More generally, the sum of two odd primes is even, and the sum of 2 and an odd prime is odd. Since 10001 is odd, the only possibility is:
\[
10001=2+p
\]
with \(p\) prime.

Then
\[
p=9999,
\]
which is not prime because it is divisible by 3.

So there are no ways.

ANSWER 12: A

Problem 1:  
We’re told that a * b means \(3a-b\). We need \(x\) if  
\[
2*(5*x)=1.
\]
First compute the inside: \(5*x = 3\cdot 5 - x = 15-x\).  
Now apply the rule again:
\[
2*(5*x)=2*(15-x)=3\cdot 2-(15-x)=6-15+x=x-9.
\]
Set equal to 1:
\[
x-9=1 \implies x=10.
\]
Check: \(5*10=3\cdot 5-10=5\), then \(2*5=6-5=1\) ✅  
ANSWER 1: D

Problem 2:  
Maria works 8 hours a day, not counting lunch. Lunch is 45 minutes.  
Total time from 7:25 A.M. to when she takes lunch at noon is:
\[
12:00 - 7:25 = 4 \text{ hours } 35 \text{ minutes}.
\]
So before lunch she works 8 hours? Careful: she only works 8 hours total in the day (excluding lunch). That means after lunch she must work the remaining part of the 8 hours.

Before lunch, from 7:25 to noon is \(4\text{ hr }35\text{ min}=4.5833\text{ hr}\).  
But she must work exactly 8 hours total excluding lunch, so after lunch she must work:
\[
8\text{ hr} - 4\text{ hr }35\text{ min} = 3\text{ hr }25\text{ min}.
\]
Lunch is 45 minutes from 12:00 to 12:45, then she works 3:25 more:
\[
12:45 + 3:25 = 4:10 \text{ P.M.}
\]
ANSWER 2: C

Problem 3:  
We want:
\[
\frac{\text{lcm}(180,594)}{\gcd(180,594)}.
\]
Use the identity for positive integers:
\[
180\cdot 594=\gcd(180,594)\cdot \text{lcm}(180,594).
\]
So:
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{(\gcd)^2}.
\]

Let’s factor:
\[
180=2^2\cdot 3^2\cdot 5,\quad
594=2\cdot 3^3\cdot 11.
\]
Thus
\[
\gcd(180,594)=2^1\cdot 3^2=2\cdot 9=18.
\]
Then
\[
\frac{\text{lcm}}{\gcd}=\frac{180\cdot 594}{18^2}.
\]
Compute:
\[
180/18=10,\quad 594/18=33.
\]
So:
\[
\frac{180\cdot 594}{18^2}=(180/18)(594/18)=10\cdot 33=330.
\]
ANSWER 3: C

Problem 4:  
Monday Taye has \(\$2\). Each day he either:
- gains \(\$3\), or
- doubles the amount.

We need the number of distinct dollar amounts possible on Thursday, which is 3 days later (Mon → Tue → Wed → Thu).

Let’s track possibilities by day.

Day 0 (Mon): \(\{2\}\)

Day 1 (Tue): from 2:
- gain: \(2+3=5\)
- double: \(2\cdot 2=4\)  
So \(\{4,5\}\)

Day 2 (Wed): from 4 and 5
- From 4: \(4+3=7\), \(4\cdot 2=8\)
- From 5: \(5+3=8\), \(5\cdot 2=10\)
So values: \(\{7,8,10\}\)

Day 3 (Thu): from 7, 8, 10
- From 7: \(7+3=10\), \(7\cdot 2=14\)
- From 8: \(8+3=11\), \(8\cdot 2=16\)
- From 10: \(10+3=13\), \(10\cdot 2=20\)

So Thursday values are \(\{10,11,13,14,16,20\}\), which is 6 values.

ANSWER 4: D

Problem 5:  
Three equally spaced integers: let them be
\[
a-d,\ a,\ a+d.
\]
First and second sum to 40:
\[
(a-d)+a=2a-d=40.
\]
Second and third sum to 60:
\[
a+(a+d)=2a+d=60.
\]
Add the equations:
\[
(2a-d)+(2a+d)=40+60 \implies 4a=100 \implies a=25.
\]
Then from \(2a-d=40\):
\[
50-d=40 \implies d=10.
\]
Numbers are \(25-10=15,\ 25,\ 25+10=35\). Sum:
\[
15+25+35=75.
\]
ANSWER 5: B

Problem 6:  
Ricardo has 2020 coins: \(p\) pennies and \(n\) nickels, with
\[
p+n=2020,\quad p\ge1,\ n\ge1.
\]
Total value in cents:
\[
V=p+5n=p+5(2020-p)=p+10100-5p=10100-4p.
\]
So value decreases as \(p\) increases.

- Greatest value occurs at smallest \(p\), i.e. \(p=1\). Then \(n=2019\):
\[
V_{\max}=1+5\cdot 2019=1+10095=10096.
\]
- Least value occurs at largest \(p\), i.e. \(n=1\Rightarrow p=2019\):
\[
V_{\min}=2019+5\cdot 1=2024.
\]
Difference:
\[
10096-2024=8072.
\]
ANSWER 6: C

Problem 7:  
In an isosceles triangle, two sides are equal, so two base angles are equal.  
Given that two angles measure \(70^\circ\) and \(x^\circ\). The possible configurations are:

1) The equal angles are \(70^\circ\) and \(70^\circ\), so \(x^\circ\) is the third angle:
\[
x=180-70-70=40.
\]
2) One \(70^\circ\) is an equal base angle and the other \(70^\circ\) is actually the remaining angle? But the problem says “two angles measure 70° and x°” meaning the triangle has angles \(\{70,70,x\}\) or \(\{70,x,x\}\) depending on which is the repeated one.

If the equal angles are \(x^\circ\) and \(x^\circ\), then the third angle is \(70^\circ\):
\[
70+ x + x = 180 \implies 2x=110 \implies x=55.
\]

So the possible values of \(x\) are \(40\) and \(55\). But the question says “three possible values.” That indicates there are three distinct ways \(x\) can be located:

Let the angles be either \((70,70,x)\) or \((70,x,x)\), but since \(x\) could correspond to either the repeated angle or the single angle, the total distinct \(x\) values across placements gives 3 values as follows:

- If \(x\) is the third angle in \((70,70,x)\): \(x=40\).
- If \(x\) is one of the base angles in \((70,x,x)\): then \(x=55\).
- The remaining case is \((x,x,70)\) which is the same as above (still \(55\)), so that doesn’t add a third.

However, an isosceles triangle has exactly two equal angles, so there are only two distinct values possible for \(x\), not three—unless the problem intends that the triangle could have angles \((70, x, x)\) with \(x\neq 70\) (giving one value), and also \((70,70,x)\) (giving another), but still that’s two.

Thus the “three possible values” must include the possibility \(x=70\) when all angles are \(70,70,40\) impossible, or when \(x=70\) gives equilateral: \(70+70+70=210\) impossible.

So we must interpret the statement as: “Two angles of an isosceles triangle measure 70° and x°, respectively,” meaning one angle is 70 and another is x, but the remaining angle could be 70 or x depending on which sides are equal. That yields:
- Equal angles are the ones labeled \(70\) and \(x\): so the triangle angles are \((70,70,x)\) with \(x=40\).
- Equal angles are the ones labeled \(x\) and the unlabeled third: angles \((70,x,x)\): \(x=55\).
- Equal angles are the ones labeled \(70\) and the unlabeled third: angles \((70,70,x)\): again \(x=40\).

Still only two values; yet the answer choices suggest a single sum of three values, which likely means the three possible \(x\) are actually \(40,50,60\) style—something is off.

Given the standard AMC8 problem: “Two angles of an isosceles triangle measure 70° and x°. What is the sum of the three possible values of x?” The three possible values are \(20,40,55\)? Let’s compute correctly by considering tha

Problem 1:
The operation is defined as $a * b = 3a - b$. We want to find $x$ such that $2 * (5 * x) = 1$.
First, evaluate $(5 * x) = 3(5) - x = 15 - x$.
Now, substitute this into the outer operation: $2 * (15 - x) = 1$.
Using the definition again: $3(2) - (15 - x) = 1$.
$6 - 15 + x = 1 \implies -9 + x = 1 \implies x = 10$.
ANSWER 1: D

Problem 2:
Maria works 8 hours plus a 45-minute lunch. Total time from start to finish is 8 hours and 45 minutes.
Start time: 7:25 A.M.
7:25 A.M. + 4 hours = 11:25 A.M.
11:25 A.M. + 35 minutes = 12:00 P.M. (noon).
Remaining time to work: 8 hours 45 minutes - 4 hours 35 minutes = 4 hours 10 minutes.
12:00 P.M. + 4 hours 10 minutes = 4:10 P.M.
ANSWER 2: C

Problem 3:
Prime factorizations: $180 = 18 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^2 \times 3^2 \times 5$ (Wait, $180 = 2^2 \times 3^2 \times 5$).
$594 = 2 \times 297 = 2 \times 3 \times 99 = 2 \times 3^3 \times 11$.
$\text{GCF}(180, 594) = 2^1 \times 3^2 = 18$.
$\text{LCM}(180, 594) = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 55 = 5940$.
Ratio = $5940 / 18 = 330$.
ANSWER 3: C

Problem 4:
Start: $M_0 = 2$.
Monday to Tuesday ($M_1$): $2+3=5$ or $2 \times 2=4$.
Tuesday to Wednesday ($M_2$):
If $M_1=5$: $5+3=8$ or $5 \times 2=10$.
If $M_1=4$: $4+3=7$ or $4 \times 2=8$. (Amounts: 7, 8, 10).
Wednesday to Thursday ($M_3$):
From 8: $8+3=11$ or $8 \times 2=16$.
From 10: $10+3=13$ or $10 \times 2=20$.
From 7: $7+3=10$ or $7 \times 2=14$.
Distinct amounts: $\{11, 16, 13, 20, 10, 14\}$. Total = 6.
ANSWER 4: D

Problem 5:
Let the integers be $a, a+d, a+2d$.
Sum 1: $a + (a+d) = 2a+d = 40$.
Sum 2: $(a+d) + (a+2d) = 2a+3d = 60$.
Subtract: $(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$.
$2a + 10 = 40 \implies 2a = 30 \implies a = 15$.
Numbers: 15, 25, 35. Sum = $15+25+35 = 75$.
ANSWER 5: B

Problem 6:
Let $p$ be pennies, $n$ be nickels. $p+n = 2020$.
Value $V = 1p + 5n = 1(2020-n) + 5n = 2020 + 4n$.
Max $n = 2019$ (since $p \ge 1$): $V_{max} = 2020 + 4(2019) = 2020 + 8076 = 10096$.
Min $n = 1$: $V_{min} = 2020 + 4(1) = 2024$.
Difference: $10096 - 2024 = 8072$.
ANSWER 6: C

Problem 7:
Case 1: Angles are 70, 70, $x$. $70+70+x=180 \implies x=40$.
Case 2: Angles are 70, $x, x$. $70+2x=180 \implies 2x=110 \implies x=55$.
Case 3: Angles are $x, 70, 70$ (same as Case 1).
Wait, the question asks for the sum of the three *possible* values. If the triangle is isosceles, the angles could be $(70, 70, 40)$ or $(70, 55, 55)$. These are only two values. Re-reading: "Two angles... measure 70 and x". If $x$ is the vertex, $x=40$. If $x$ is a base angle, $x=70$ (already counted) or $x=55$. The possible values for $x$ are 40, 55, 70. Sum = $40+55+70 = 165$.
ANSWER 7: D

Problem 8:
$2 / (1 - 2/3) = 2 / (1/3) = 2 \times 3 = 6$.
ANSWER 8: E

Problem 9:
Number $N$ satisfies $N \equiv 2 \pmod 3, 4, 5, 6$.
$N-2$ is a multiple of $\text{LCM}(3, 4, 5, 6) = 60$.
$N = 60k + 2$.
For $k=1, N=62$. For $k=2, N=122$. For $k=3, N=182$. For $k=4, N=242$. For $k=5, N=302$.
The smallest such number is 62, which lies between 60 and 79.
ANSWER 9: B

Problem 10:
Digits $\{d_1, d_2, d_3\}$ such that $d_1 d_2 d_3 = 24$.
Factors of 24: 1, 2, 3, 4, 6, 8.
Sets: $\{1, 3, 8\} \to 6$ perms; $\{1, 4, 6\} \to 6$ perms; $\{2, 2, 6\} \to 3$ perms; $\{2, 3, 4\} \to 6$ perms.
Total = $6+6+3+6 = 21$.
ANSWER 10: D

Problem 11:
Ben: $6 \to (6+1) \times 2 = 14$.
Sue: $14 \to (14-1) \times 2 = 13 \times 2 = 26$.
ANSWER 11: C

Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is 2.
$10001 = 2 + p \implies p = 9999$.
9999 is divisible by 3 ($9+9+9+9=36$), so it is not prime.
There are no such pairs.
ANSWER 12: A

Problem 1:
The operation is defined by \(a * b = 3a - b\). First evaluate the inner operation:  
\(5 * x = 3 \cdot 5 - x = 15 - x\).  
Now apply the outer operation:  
\(2 * (15 - x) = 3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9\).  
Set the result equal to 1:  
\(x - 9 = 1 \implies x = 10\).  
This matches choice D (and rules out the other listed values).

ANSWER 1: D

Problem 2:
Maria must work a total of 8 hours. The interval from 7:25 A.M. to noon is 4 hours 35 minutes, so the remaining work time needed is  
\(8\) hours \(- 4\) hours \(35\) minutes \(= 3\) hours \(25\) minutes.  
Adding the 45-minute lunch break that begins at noon gives a finish time of 12:45 P.M. + 3 hours 25 minutes = 4:10 P.M.  
(Equivalently, the total elapsed time from start to finish is 8 hours 45 minutes, and 7:25 A.M. + 8 hours 45 minutes is also 4:10 P.M.)

ANSWER 2: C

Problem 3:
Prime factorizations: \(180 = 2^2 \cdot 3^2 \cdot 5\) and \(594 = 2 \cdot 3^3 \cdot 11\).  
Thus \(\operatorname{GCF}(180,594) = 2 \cdot 3^2 = 18\) and  
\(\operatorname{LCM}(180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940\).  
The required ratio is \(5940 / 18 = 330\).

ANSWER 3: C

Problem 4:
Three successive operations (each either “add 3” or “double”) are applied to the starting amount 2. The eight possible sequences produce the final amounts  
11, 16, 13, 20, 10, 14, 11, 16.  
The distinct values are 10, 11, 13, 14, 16, 20—exactly six different dollar amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be \(a\), \(a+d\), \(a+2d\). The given sums become the system  
\(2a + d = 40\),  
\(2a + 3d = 60\).  
Subtracting yields \(2d = 20\), so \(d = 10\). Substituting gives \(2a = 30\), hence \(a = 15\).  
The integers are therefore 15, 25, 35 whose sum is 75.

ANSWER 5: B

Problem 6:
The extreme values occur when the number of nickels is maximized or minimized (subject to at least one penny and one nickel).  
Maximum value: 2019 nickels and 1 penny give \(2019 \cdot 5 + 1 = 10096\) cents.  
Minimum value: 1 nickel and 2019 pennies give \(5 + 2019 = 2024\) cents.  
The difference is \(10096 - 2024 = 8072\) cents.

ANSWER 6: C

Problem 7:
An isosceles triangle has angles 70°, \(x^\circ\), and \(110 - x^\circ\). Two of the three angles must be equal, which produces the three admissible cases  
\(x = 70\) (third angle 40°),  
\(x = 40\) (third angle 70°),  
\(x = 55\) (third angle 55°).  
Their sum is \(70 + 40 + 55 = 165\).

ANSWER 7: D

Problem 8:
Simplify the denominator first:  
\(1 - 2/3 = 1/3\).  
Division then yields  
\(2 \div (1/3) = 2 \cdot 3 = 6\).

ANSWER 8: E

Problem 9:
The stated remainders imply that \(n-2\) is divisible by each of 3, 4, 5, 6, hence by their least common multiple 60.  
Thus \(n = 60k + 2\). The smallest integer greater than 2 occurs when \(k=1\), giving \(n=62\), which lies between 60 and 79.

ANSWER 9: B

Problem 10:
All ordered triples of digits \((a,b,c)\) with \(a \in \{1,\dots,9\}\) and \(b,c \in \{0,\dots,9\}\) whose product equals 24 must be counted. The admissible unordered triples and the number of distinct permutations that keep the first digit nonzero are  
\{1,3,8\} (6 permutations),  
\{1,4,6\} (6 permutations),  
\{2,2,6\} (3 permutations),  
\{2,3,4\} (6 permutations).  
The total is \(6+6+3+6=21\).

ANSWER 10: D

Problem 11:
Ben starts with 6. Adding 1 and doubling produces 14. Sue then subtracts 1 and doubles:  
\(14-1=13\), \(13 \cdot 2 = 26\).

ANSWER 11: C

Problem 12:
Any sum of two primes equaling the odd integer 10001 must consist of the even prime 2 together with the odd integer 9999. But 9999 is divisible by 3 (digit sum 36), hence composite. Consequently no such pair of primes exists.

ANSWER 12: A

Problem 1:
We need to find the value of x if 2 ∗ (5 ∗ x) = 1, where a ∗ b means 3a - b.
First, let's evaluate 5 ∗ x. Using the definition of ∗, we have 5 ∗ x = 3(5) - x = 15 - x.
Now, we need to evaluate 2 ∗ (15 - x). Again, using the definition of ∗, we have 2 ∗ (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.
We are given that 2 ∗ (5 ∗ x) = 1, so x - 9 = 1.
Solving for x, we get x = 1 + 9 = 10.
Let's check the answer choices to see if any of them are tempting wrong choices. 
A. 1/10 is too small.
B. 2 is too small.
C. 10/3 is not equal to 10.
D. 10 matches our solution.
E. 14 is too large.
Thus, our solution is correct.

ANSWER 1: D

Problem 2:
Maria works for 8 hours and takes a 45-minute lunch break. She starts working at 7:25 A.M. and takes her lunch break at noon.
First, let's calculate the time she spends working before lunch. From 7:25 A.M. to noon is 4 hours and 35 minutes.
Since she works for 8 hours in total, she needs to work for 8 hours - 4 hours 35 minutes = 3 hours 25 minutes after lunch.
She takes her lunch break at noon and it lasts for 45 minutes, so she resumes work at 12:45 P.M.
Adding 3 hours and 25 minutes to 12:45 P.M., we get 4:10 P.M.
Let's check the answer choices.
A. 3:40 P.M. is too early.
B. 3:55 P.M. is too early.
C. 4:10 P.M. matches our solution.
D. 4:25 P.M. is too late.
E. 4:40 P.M. is too late.
Thus, our solution is correct.

ANSWER 2: C

Problem 3:
We need to find the ratio of the least common multiple (LCM) of 180 and 594 to the greatest common factor (GCF) of 180 and 594.
First, let's find the prime factorization of 180 and 594.
180 = 2^2 * 3^2 * 5
594 = 2 * 3^3 * 11
The GCF is the product of the common prime factors raised to the lowest power.
GCF = 2 * 3^2 = 18
The LCM is the product of the highest powers of all prime factors.
LCM = 2^2 * 3^3 * 5 * 11 = 5940
Now, we can find the ratio of LCM to GCF.
Ratio = LCM / GCF = 5940 / 18 = 330
Let's check the answer choices.
A. 110 is too small.
B. 165 is too small.
C. 330 matches our solution.
D. 625 is too large.
E. 660 is too large.
Thus, our solution is correct.

ANSWER 3: C

Problem 4:
Taye starts with $2 on Monday and either gains $3 or doubles his money each day.
Let's analyze the possible amounts of money he can have on Tuesday, Wednesday, and Thursday.
On Tuesday, he can have $2 + $3 = $5 or $2 * 2 = $4.
On Wednesday, if he had $5 on Tuesday, he can have $5 + $3 = $8 or $5 * 2 = $10. If he had $4 on Tuesday, he can have $4 + $3 = $7 or $4 * 2 = $8.
So, on Wednesday, he can have $7, $8, or $10.
On Thursday, if he had $7 on Wednesday, he can have $7 + $3 = $10 or $7 * 2 = $14. If he had $8 on Wednesday, he can have $8 + $3 = $11 or $8 * 2 = $16. If he had $10 on Wednesday, he can have $10 + $3 = $13 or $10 * 2 = $20.
So, on Thursday, he can have $10, $11, $13, $14, $16, or $20.
There are 6 different possible amounts of money Taye can have on Thursday, but we should list them out to be sure: $10, $11, $13, $14, $16, $20.
Let's check the answer choices.
A. 3 is too small.
B. 4 is too small.
C. 5 is too small.
D. 6 matches our solution.
E. 7 is too large.
Thus, our solution is correct, but we should note that we directly counted the possibilities.

ANSWER 4: D

Problem 5:
Jami picked three equally spaced integers on the number line. Let's call them x, x + d, and x + 2d, where d is the common difference.
We are given that x + (x + d) = 40 and (x + d) + (x + 2d) = 60.
Simplifying the first equation, we get 2x + d = 40.
Simplifying the second equation, we get 2x + 3d = 60.
Subtracting the first equation from the second, we get 2d = 20, so d = 10.
Substituting d = 10 into the first equation, we get 2x + 10 = 40, so 2x = 30 and x = 15.
Now, we can find the three numbers: x = 15, x + d = 25, and x + 2d = 35.
The sum of the three numbers is 15 + 25 + 35 = 75.
Let's check the answer choices.
A. 70 is too small.
B. 75 matches our solution.
C. 80 is too large.
D. 85 is too large.
E. 90 is too large.
Thus, our solution is correct.

ANSWER 5: B

Problem 6:
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel.
To maximize the amount of money, he should have as many nickels as possible. Since he has at least one penny, he can have 2019 nickels and 1 penny.
The maximum amount of money is 2019 * 5 cents + 1 * 1 cent = 10095 cents + 1 cent = 10096 cents.
To minimize the amount of money, he should have as many pennies as possible. Since he has at least one nickel, he can have 2019 pennies and 1 nickel.
The minimum amount of money is 2019 * 1 cent + 1 * 5 cents = 2019 cents + 5 cents = 2024 cents.
The difference between the maximum and minimum amounts of money is 10096 cents - 2024 cents = 8072 cents.
Let's check the answer choices.
A. 8062 is too small.
B. 8068 is too small.
C. 8072 matches our solution.
D. 8076 is too large.
E. 8082 is too large.
Thus, our solution is correct.

ANSWER 6: C

Problem 7:
Two angles of an isosceles triangle measure 70° and x°.
Since it's an isosceles triangle, two angles must be equal. There are three possibilities:
1. The two equal angles are 70°, so x = 70° and the third angle is 180° - 70° - 70° = 40°.
2. The two equal angles are x°, so 70° + x + x = 180°, which gives 2x = 110° and x = 55°.
3. One of the equal angles is 70° and the other is x, so x = 70° (already considered in case 1) or 70° = x, which is the same as case 1. The other possibility is that x is the vertex angle, so x + 70° + 70° = 180°, giving x = 40°.
The three possible values of x are 70°, 55°, and 40°.
The sum of these values is 70° + 55° + 40° = 165°.
Let's check the answer choices.
A. 95 is too small.
B. 125 is too small.
C. 140 is too small.
D. 165 matches our solution.
E. 180 is too large.
Thus, our solution is correct.

ANSWER 7: D

Problem 8:
We need to simplify the expression 2 / (1 - 2/3).
First, let's simplify the denominator: 1 - 2/3 = (3/3) - (2/3) = 1/3.
Now, we can rewrite the expression as 2 / (1

**Problem 1:**  
Given: \(a * b = 3a - b\). We need \(x\) such that \(2 * (5 * x) = 1\).  
First, evaluate \(5 * x\): \(5 * x = 3(5) - x = 15 - x\).  
Next, \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).  
Set equal to 1: \(x - 9 = 1 \implies x = 10\).  
Answer: 10, which is choice D.

ANSWER 1: D

**Problem 2:**  
Maria works 8 hours, excluding 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
Time from 7:25 A.M. to noon: 7:25–8:00 is 35 min, 8:00–12:00 is 4 h, total = 4 h 35 min.  
Remaining work time = 8 h – 4 h 35 min = 3 h 25 min.  
After noon, she takes 45 min lunch then works 3 h 25 min, so end time = noon + 45 min + 3 h 25 min = noon + 4 h 10 min = 4:10 P.M.  
Answer: 4:10 P.M., choice C.

ANSWER 2: C

**Problem 3:**  
Find ratio LCM(180, 594) / GCF(180, 594).  
Prime factorizations:  
\(180 = 2^2 \cdot 3^2 \cdot 5^1\)  
\(594 = 2^1 \cdot 3^3 \cdot 11^1\)  
GCF = \(2^{\min(2,1)} \cdot 3^{\min(2,3)} = 2^1 \cdot 3^2 = 18\).  
LCM = \(2^2 \cdot 3^3 \cdot 5^1 \cdot 11^1 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940\).  
Ratio = \(5940 / 18 = 330\).  
Answer: 330, choice C.

ANSWER 3: C

**Problem 4:**  
Start Monday: $2. Each day: either add $3 or double.  
Tuesday: \(2+3=5\) or \(2\times2=4\) → amounts: $4, $5.  
Wednesday from $4: \(4+3=7\), \(4\times2=8\); from $5: \(5+3=8\), \(5\times2=10\). Unique: $7, $8, $10.  
Thursday: from $7 → 10, 14; from $8 → 11, 16; from $10 → 13, 20.  
Unique amounts: $10, $11, $13, $14, $16, $20. That is 6 different amounts.  
Answer: 6, choice D.

ANSWER 4: D

**Problem 5:**  
Three equally spaced integers: \(a, a+d, a+2d\).  
Given: \(a + (a+d) = 40\) → \(2a + d = 40\).  
\((a+d) + (a+2d) = 60\) → \(2a + 3d = 60\).  
Subtract first from second: \(2d = 20 \Rightarrow d = 10\). Then \(2a + 10 = 40 \Rightarrow a = 15\).  
Numbers: 15, 25, 35. Sum = 75.  
Answer: 75, choice B.

ANSWER 5: B

**Problem 6:**  
Let \(p\) = pennies, \(n\) = nickels. \(p + n = 2020\), \(p,n \ge 1\).  
Total cents = \(p + 5n = (2020 - n) + 5n = 2020 + 4n\).  
Maximum occurs at max \(n = 2019\): cents = \(2020 + 4(2019) = 2020 + 8076 = 10096\).  
Minimum occurs at min \(n = 1\): cents = \(2020 + 4(1) = 2024\).  
Difference = \(10096 - 2024 = 8072\).  
Answer: 8072, choice C.

ANSWER 6: C

**Problem 7:**  
Isosceles triangle with angles \(70^\circ\) and \(x^\circ\). Two angles are equal. Three possibilities:  
- The two \(70^\circ\) angles are equal: third angle = \(180 - 2(70) = 40^\circ \Rightarrow x = 40\).  
- The two \(x^\circ\) angles are equal: \(2x + 70 = 180 \Rightarrow 2x = 110 \Rightarrow x = 55\).  
- One \(70^\circ\) and one \(x^\circ\) are the equal pair: then \(x = 70\) (and triangle has angles 70, 70, 40).  
Possible values: 40, 55, 70. Sum = \(40+55+70 = 165\).  
Answer: 165, choice D.

ANSWER 7: D

**Problem 8:**  
\( \frac{2}{\,1 - \frac{2}{3}\,} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6\).  
Answer: 6, choice E.

ANSWER 8: E

**Problem 9:**  
Number \(n > 2\) leaves remainder 2 when divided by 3, 4, 5, 6.  
So \(n - 2\) is a multiple of LCM(3,4,5,6).  
LCM(3,4,5,6) = LCM(4,5,6) = 60.  
Smallest \(n > 2\): \(n - 2 = 60 \Rightarrow n = 62\).  
62 is between 60 and 79.  
Answer: B.

ANSWER 9: B

**Problem 10:**  
Three-digit numbers (digits from 1–9, product = 24).  
List digit triples (a,b,c) with product 24:  
\{1,3,8\}, \{1,4,6\}, \{2,2,6\}, \{2,3,4\}.  
Permutations:  
\{1,3,8\}: 3! = 6  
\{1,4,6\}: 3! = 6  
\{2,3,4\}: 3! = 6  
\{2,2,6\}: 3!/2! = 3  
Total = \(6+6+6+3 = 21\).  
Answer: 21, choice D.

ANSWER 10: D

**Problem 11:**  
Ben thinks 6 → add 1 = 7 → double = 14 → gives 14 to Sue.  
Sue: subtract 1: 14 – 1 = 13 → double: 26.  
Answer: 26, choice C.

ANSWER 11: C

**Problem 12:**  
10001 is odd. Sum of two primes: must be even + odd. Only even prime is 2.  
If one prime = 2, the other = \(10001 - 2 = 9999\).  
Check if 9999 is prime: \(9999 = 9 \times 1111 = 3^2 \times 11 \times 101\), composite.  
No other possibilities (odd+odd = even). So 0 ways.  
Answer: 0, choice A.

ANSWER 12: A

Problem 1:
We are told that $a * b = 3a - b$. We need to find $x$ such that $2 * (5 * x) = 1$.

**Step 1: Evaluate the inner operation $5 * x$.**
Using the definition: $5 * x = 3(5) - x = 15 - x$.

**Step 2: Evaluate the outer operation $2 * (15 - x)$.**
$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9$.

**Step 3: Set equal to 1 and solve.**
$x - 9 = 1 \implies x = 10$.

Checking: $5 * 10 = 15 - 10 = 5$, then $2 * 5 = 6 - 5 = 1$. ✓

ANSWER 1: D

---

Problem 2:
Maria works 8 hours total (not counting her 45-minute lunch). She starts at 7:25 AM and takes lunch at noon.

**Step 1: Calculate morning work time.**
From 7:25 AM to 12:00 PM (noon) = 4 hours 35 minutes.

**Step 2: Calculate remaining work time needed.**
$8\text{ hours} - 4\text{ hours } 35\text{ minutes} = 3\text{ hours } 25\text{ minutes}$.

**Step 3: Determine when lunch ends.**
Lunch starts at noon and lasts 45 minutes, so lunch ends at 12:45 PM.

**Step 4: Add remaining work time.**
$12:45 \text{ PM} + 3\text{ hours } 25\text{ minutes} = 4:10 \text{ PM}$.

ANSWER 2: C

---

Problem 3:
We need $\frac{\text{LCM}(180, 594)}{\text{GCF}(180, 594)}$.

**Step 1: Prime factorizations.**
$180 = 2^2 \times 3^2 \times 5$
$594 = 2 \times 3^3 \times 11$

**Step 2: Find GCF.**
$\text{GCF} = 2^1 \times 3^2 = 18$

**Step 3: Find LCM.**
$\text{LCM} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$

**Step 4: Compute the ratio.**
$\frac{5940}{18} = 330$

(Alternatively, $\frac{\text{LCM}}{\text{GCF}} = \frac{a \times b}{\text{GCF}^2} = \frac{180 \times 594}{18^2} = \frac{106920}{324} = 330$.)

ANSWER 3: C

---

Problem 4:
Taye starts with \$2 on Monday. Each of the next 3 days (Tue, Wed, Thu), he either adds \$3 or doubles. We track all paths.

**Monday:** 2

**Tuesday:** $2+3=5$ or $2\times2=4$

**Wednesday:**
- From 5: $5+3=8$ or $5\times2=10$
- From 4: $4+3=7$ or $4\times2=8$

**Thursday:**
- From 8: $8+3=11$ or $8\times2=16$
- From 10: $10+3=13$ or $10\times2=20$
- From 7: $7+3=10$ or $7\times2=14$
- From 8: $8+3=11$ or $8\times2=16$

**Collecting all Thursday values:** $\{11, 16, 13, 20, 10, 14, 11, 16\}$

**Unique values:** $\{10, 11, 13, 14, 16, 20\}$ — that's **6** different amounts.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers: $a,\; a+d,\; a+2d$.

**Set up equations:**
- First + Second: $a + (a+d) = 2a + d = 40$
- Second + Third: $(a+d) + (a+2d) = 2a + 3d = 60$

**Subtract the first from the second:**
$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10$

**Solve for $a$:**
$2a + 10 = 40 \implies a = 15$

**The three numbers:** $15, 25, 35$

**Sum:** $15 + 25 + 35 = 75$

ANSWER 5: B

---

Problem 6:
Ricardo has 2020 coins (pennies and nickels), with $p \geq 1$ and $n \geq 1$, and $p + n = 2020$.

**Total value:** $V = p + 5n = p + 5(2020 - p) = 10100 - 4p$

**Greatest value:** Minimize $p$. Since $p \geq 1$, set $p = 1$:
$V_{\max} = 10100 - 4(1) = 10096$

**Least value:** Maximize $p$. Since $n \geq 1$, set $p = 2019$:
$V_{\min} = 10100 - 4(2019) = 10100 - 8076 = 2024$

**Difference:** $10096 - 2024 = 8072$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70°$ and $x°$. At least two angles must be equal.

**Case 1: The two equal angles are both $70°$.**
Then $x$ could be the third angle: $x = 180 - 70 - 70 = 40$.
(Or $x$ could be one of the $70°$ angles, giving $x = 70$.)

**Case 2: The two equal angles are both $x°$, and $70°$ is the odd angle.**
$2x + 70 = 180 \implies 2x = 110 \implies x = 55$.

**Case 3: One equal angle is $70°$ and the other is $x°$.**
This means $x = 70$, giving the triangle $(70, 70, 40)$, which we already found.

The three possible values of $x$ are: $40, 55, 70$.

**Sum:** $40 + 55 + 70 = 165$

ANSWER 7: D

---

Problem 8:
Evaluate $\frac{2}{1 - \frac{2}{3}}$.

**Step 1: Simplify the denominator.**
$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

**Step 2: Divide.**
$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6$

ANSWER 8: E

---

Problem 9:
We need the smallest whole number $N > 2$ such that $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{6}$.

This means $N - 2$ is divisible by 3, 4, 5, and 6.

**Find LCM(3, 4, 5, 6):**
$3 = 3,\quad 4 = 2^2,\quad 5 = 5,\quad 6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 60$

So $N - 2 = 60k$, meaning $N = 60k + 2$.

The smallest $N > 2$ occurs at $k = 1$: $N = 62$.

$62$ lies between **60 and 79**.

ANSWER 9: B

---

Problem 10:
We need 3-digit numbers whose digits multiply to 24. Find all unordered triples of digits $(a, b, c)$ with $a \leq b \leq c$ and $abc = 24$:

| Triple | Permutations |
|--------|-------------|
| $(1, 3, 8)$ | $3! = 6$ |
| $(1, 4, 6)$ | $3! = 6$ |
| $(2, 2, 6)$ | $3!/2! = 3$ |
| $(2, 3, 4)$ | $3! = 6$ |

(Other factorizations like $1 \times 1 \times 24$ or $1 \times 2 \times 12$ use numbers $> 9$, so they're invalid.)

**Total:** $6 + 6 + 3 + 6 = 21$

ANSWER 10: D

---

Problem 11:
**Ben's turn:** Ben thinks of 6.
- Add 1: $6 + 1 = 7$
- Double: $7 \times 2 = 14$
- Ben gives **14** to Sue.

**Sue's turn:** Sue receives 14.
- Subtract 1: $14 - 1 = 13$
- Double: $13 \times 2 = 26$

Sue's answer is **26**.

ANSWER 11: C

---

Problem 12:
We need $p_1 + p_2 = 10001$ where both are primes.

Since 10001 is **odd**, the sum of two primes can only be odd if one of them is **2** (the only even prime). So we need $2 + p = 10001$, giving $p = 9999$.

**Check if 9999 is prime:**
$9999 = 9 \times 1111 = 9 \times 11 \times 101$

Since 9999 is divisible by 3 (and 9, 11, 101), it is **not prime**.

Therefore, there are **0** ways to write 10001 as the sum of two primes.

ANSWER 12: A

 Problem 1:
We are given a custom operation defined by $a * b = 3a - b$, and we must solve for $x$ in the equation $2 * (5 * x) = 1$.

First evaluate the inner expression $5 * x$:
$$5 * x = 3(5) - x = 15 - x.$$

Now substitute this into the outer operation:
$$2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9.$$

Set this equal to $1$:
$$x - 9 = 1 \implies x = 10.$$

We can check: $5 * 10 = 15 - 10 = 5$, and $2 * 5 = 6 - 5 = 1$. This matches. Among the choices, $10$ is option D.

ANSWER 1: D

---

Problem 2:
Maria must work $8$ hours excluding a $45$-minute lunch. She starts at $7\!:\!25$ A.M. and begins lunch at noon. We need her ending time.

Time worked before lunch:
From $7\!:\!25$ A.M. to $12\!:\!00$ P.M. is $4$ hours $35$ minutes.

Remaining work time:
$$8\text{ h} - 4\text{ h }35\text{ min} = 3\text{ h }25\text{ min}.$$

She returns from lunch at $12\!:\!45$ P.M. Adding the remaining $3$ hours $25$ minutes:
$$12\!:\!45 + 3\!:\!25 = 4\!:\!10\text{ P.M.}$$

(Equivalently, the whole span from start to finish is $8$ hours of work plus $45$ minutes lunch, totaling $8$ h $45$ min; $7\!:\!25$ A.M. $+$ $8\!:\!45$ $=$ $4\!:\!10$ P.M.)

ANSWER 2: C

---

Problem 3:
We need the ratio $\dfrac{\text{lcm}(180,594)}{\text{gcd}(180,594)}$.

Prime factorizations:
$$180 = 2^2 \cdot 3^2 \cdot 5,$$
$$594 = 2 \cdot 3^3 \cdot 11.$$

Greatest common factor (take minimum exponents):
$$\gcd = 2^1 \cdot 3^2 = 2 \cdot 9 = 18.$$

Least common multiple (take maximum exponents):
$$\text{lcm} = 2^2 \cdot 3^3 \cdot 5 \cdot 11 = 4 \cdot 27 \cdot 5 \cdot 11 = 5940.$$

The desired ratio is:
$$\frac{5940}{18} = 330.$$

ANSWER 3: C

---

Problem 4:
Taye starts with $\$2$ on Monday. On each of the next three days (Tuesday, Wednesday, Thursday) he either gains $\$3$ or doubles his money. We count the distinct amounts he could have on Thursday.

Build the tree of possibilities:
- **Tuesday:** $2+3=5$ or $2\cdot2=4$.
- **Wednesday:**  
  from $5$: $5+3=8$, $5\cdot2=10$;  
  from $4$: $4+3=7$, $4\cdot2=8$.  
  Distinct amounts: $\{7,8,10\}$.
- **Thursday:**  
  from $7$: $7+3=10$, $7\cdot2=14$;  
  from $8$: $8+3=11$, $8\cdot2=16$;  
  from $10$: $10+3=13$, $10\cdot2=20$.  

The distinct Thursday amounts are $\{10,11,13,14,16,20\}$, which is $6$ different values.

ANSWER 4: D

---

Problem 5:
Three equally spaced integers can be written as $a$, $a+d$, $a+2d$.

Given:
$$a + (a+d) = 40 \implies 2a + d = 40,$$
$$(a+d) + (a+2d) = 60 \implies 2a + 3d = 60.$$

Subtract the first equation from the second:
$$(2a+3d) - (2a+d) = 60 - 40 \implies 2d = 20 \implies d = 10.$$

Then $2a + 10 = 40 \implies a = 15$.

The three integers are $15$, $25$, $35$. Their sum is:
$$15 + 25 + 35 = 75.$$

(Shortcut: adding the two given sums yields $4a+4d=100$, so the middle term $a+d=25$, and the total sum is $3\cdot25=75$.)

ANSWER 5: B

---

Problem 6:
Ricardo has $2020$ coins, at least one penny and at least one nickel. We need the difference between the greatest and least possible total values in cents.

- **Greatest amount:** maximize nickels. Use $2019$ nickels and $1$ penny.
  $$2019 \times 5 + 1 = 10095 + 1 = 10096\text{ cents}.$$
- **Least amount:** minimize nickels. Use $1$ nickel and $2019$ pennies.
  $$5 + 2019 = 2024\text{ cents}.$$

Difference:
$$10096 - 2024 = 8072\text{ cents}.$$

ANSWER 6: C

---

Problem 7:
An isosceles triangle has angles $70^\circ$, $x^\circ$, and a third angle. Because it is isosceles, two angles must be equal. We consider the three distinct cases for $x$:

1. **The two equal angles are the base angles $70^\circ$ and $70^\circ$.**  
   Then $x$ is the vertex angle: $x = 180 - 2(70) = 40$.

2. **The two equal angles are the base angles $x$ and $x$.**  
   Then $70 + 2x = 180 \implies 2x = 110 \implies x = 55$.

3. **The two equal angles are $70^\circ$ and $x^\circ$.**  
   Then $x = 70$, and the third angle is $180 - 2(70) = 40$. This is a valid isosceles triangle.

The three possible values of $x$ are $40$, $55$, and $70$. Their sum is:
$$40 + 55 + 70 = 165.$$

ANSWER 7: D

---

Problem 8:
Evaluate
$$\frac{2}{1 - \frac{2}{3}}.$$

First simplify the denominator:
$$1 - \frac{2}{3} = \frac{1}{3}.$$

Now divide:
$$\frac{2}{\frac{1}{3}} = 2 \times 3 = 6.$$

ANSWER 8: E

---

Problem 9:
We seek the smallest whole number $N>2$ such that $N$ leaves remainder $2$ when divided by $3$, $4$, $5$, and $6$.

This means $N-2$ is divisible by each of $3,4,5,6$. Hence $N-2$ is a multiple of $\text{lcm}(3,4,5,6)$.

Compute the LCM:
$$\text{lcm}(3,4,5,6) = \text{lcm}(2^2, 3, 5) = 60.$$

So $N = 60k + 2$ for some positive integer $k$. The smallest such $N>2$ is when $k=1$:
$$N = 62.$$

The number $62$ lies between $60$ and $79$.

ANSWER 9: B

---

Problem 10:
We need the count of $3$-digit positive integers with digits $a,b,c$ (each from $1$ to $9$, since the product is $24\neq 0$) such that $abc = 24$.

Find unordered triples of digits whose product is $24$:
- $\{1,3,8\}$
- $\{1,4,6\}$
- $\{2,2,6\}$
- $\{2,3,4\}$

Count distinct permutations (ordered triples) for each:
- $\{1,3,8\}$: $3! = 6$ numbers.
- $\{1,4,6\}$: $3! = 6$ numbers.
- $\{2,2,6\}$: $\frac{3!}{2!} = 3$ numbers.
- $\{2,3,4\}$: $3! = 6$ numbers.

Total:
$$6 + 6 + 3 + 6 = 21.$$

ANSWER 10: D

---

Problem 11:
Ben thinks of $6$. Following the teacher's rule:
- Add $1$: $6+1=7$.
- Double: $7 \times 2 = 14$. Ben gives $14$ to Sue.

Sue now follows her rule:
- Subtract $1$: $14-1=13$.
- Double: $13 \times 2 = 26$.

Thus Sue's answer is $26$.

ANSWER 11: C

---

Problem 12:
We want to write $10001$ as a sum of two primes: $p + q = 10001$.

Since $10001$ is odd, one of the primes must be even (because the sum of two odd numbers is even). The only even prime is $2$.

So the only possible representation is $2 + 9999$. But $9999$ is divisible by $3$ (since $9+9+9+9=36$), and indeed $9999 = 3^2 \times 11 \times 101$, so it is not prime.

Therefore there are $0$ ways.

ANSWER 12: A

Problem 1:
We need to find x such that 2 ∗ (5 ∗ x) = 1, where a ∗ b = 3a − b.

First, compute 5 ∗ x:
5 ∗ x = 3(5) − x = 15 − x

Now compute 2 ∗ (15 − x):
2 ∗ (15 − x) = 3(2) − (15 − x) = 6 − 15 + x = x − 9

Set equal to 1:
x − 9 = 1
x = 10

ANSWER 1: D

Problem 2:
Maria works 8 hours total (not including 45-minute lunch). She starts at 7:25 A.M. and takes lunch at noon.

Work before lunch: 7:25 A.M. to noon = 4 hours 35 minutes

Remaining work after lunch: 8 hours − 4 hours 35 minutes = 3 hours 25 minutes

She returns from lunch at 12:45 P.M. Adding 3 hours 25 minutes:
12:45 P.M. + 3:25 = 4:10 P.M.

ANSWER 2: C

Problem 3:
Find LCM(180, 594) / GCF(180, 594).

Prime factorizations:
180 = 2² × 3² × 5
594 = 2 × 3³ × 11

GCF = 2¹ × 3² = 18
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 55 = 5940

Ratio = 5940 / 18 = 330

ANSWER 3: C

Problem 4:
Monday: $2

Tuesday: 2+3=5 or 2×2=4 → {4, 5}

Wednesday:
From 4: 4+3=7 or 4×2=8
From 5: 5+3=8 or 5×2=10
→ {7, 8, 10}

Thursday:
From 7: 7+3=10 or 7×2=14
From 8: 8+3=11 or 8×2=16
From 10: 10+3=13 or 10×2=20
→ {10, 11, 13, 14, 16, 20}

That's 6 different amounts.

ANSWER 4: D

Problem 5:
Let the three equally spaced integers be a, a+d, a+2d.

a + (a+d) = 40 → 2a + d = 40
(a+d) + (a+2d) = 60 → 2a + 3d = 60

Subtracting: 2d = 20, so d = 10
Then 2a + 10 = 40, so a = 15

Numbers: 15, 25, 35. Sum = 75.

ANSWER 5: B

Problem 6:
Let p = pennies, n = nickels. p + n = 2020, p ≥ 1, n ≥ 1.

Amount = p + 5n = p + 5(2020 − p) = 10100 − 4p

Maximum: p = 1 → 10100 − 4 = 10096 cents
Minimum: p = 2019 → 10100 − 8076 = 2024 cents

Difference = 10096 − 2024 = 8072

ANSWER 6: C

Problem 7:
For an isosceles triangle with angles 70° and x°:

Case 1: Two equal angles are both 70° → x = 70 (third angle = 40°)
Case 2: x equals the third angle → 70 + 2x = 180 → x = 55
Case 3: 70° equals the third angle (not x) → 2(70) + x = 180 → x = 40

Sum of possible values = 70 + 55 + 40 = 165

ANSWER 7: D

Problem 8:
2 / (1 − 2/3) = 2 / (1/3) = 2 × 3 = 6

ANSWER 8: E

Problem 9:
The number n leaves remainder 2 when divided by 3, 4, 5, and 6.
So n − 2 is divisible by all of 3, 4, 5, 6, meaning n − 2 is a multiple of LCM(3,4,5,6) = 60.

n = 60k + 2. Smallest value > 2 is when k = 1: n = 62.

62 lies between 60 and 79.

ANSWER 9: B

Problem 10:
Factorizations of 24 into three digits (1-9):
- 1 × 3 × 8: 3! = 6 arrangements
- 1 × 4 × 6: 3! = 6 arrangements
- 2 × 2 × 6: 3!/2! = 3 arrangements
- 2 × 3 × 4: 3! = 6 arrangements

Total = 6 + 6 + 3 + 6 = 21

ANSWER 10: D

Problem 11:
Ben thinks of 6. Add 1: 6 + 1 = 7. Double: 7 × 2 = 14. Ben gives 14 to Sue.

Sue subtracts 1: 14 − 1 = 13. Double: 13 × 2 = 26.

ANSWER 11: C

Problem 12:
10001 is odd. For the sum of two primes to be odd, one must be 2 (the only even prime).
So we need 2 + p = 10001, giving p = 9999.

Check if 9999 is prime: 9999 = 9 × 1111 = 9 × 11 × 101 = 99 × 101. Not prime.

Therefore, there are 0 ways.

ANSWER 12: A

Problem 1:
The operation is defined by \(a \ast b = 3a - b\).

First evaluate the inner expression \(5 \ast x\):
\[
5 \ast x = 3\cdot5 - x = 15 - x.
\]

Now evaluate the outer operation:
\[
2 \ast (5 \ast x) = 2 \ast (15 - x) = 3\cdot2 - (15 - x) = 6 - 15 + x = x - 9.
\]

Set this equal to the given value:
\[
x - 9 = 1 \quad\Longrightarrow\quad x = 10.
\]

ANSWER 1: D

---

Problem 2:
Maria must work 8 hours (480 minutes) exclusive of her 45‑minute lunch.

She starts at 7:25 AM and works until noon:
\[
\text{Work before lunch}=12{:}00 - 7{:}25 = 4\text{ h }35\text{ m}=275\text{ min}.
\]

Work remaining after lunch:
\[
480 - 275 = 205\text{ min}=3\text{ h }25\text{ m}.
\]

She finishes lunch at 12:45 PM, then works another 3 h 25 m, ending at
\[
12{:}45 + 3{:}25 = 4{:}10\text{ PM}.
\]

ANSWER 2: C

---

Problem 3:
Factor the numbers.

\[
180 = 2^{2}\cdot3^{2}\cdot5,\qquad
594 = 2\cdot3^{3}\cdot11.
\]

Greatest common factor (GCF):
\[
\text{gcd}=2^{\min(2,1)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot11^{\min(0,1)}
=2^{1}\cdot3^{2}=18.
\]

Least common multiple (LCM):
\[
\text{lcm}=2^{\max(2,1)}\cdot3^{\max(2,3)}\cdot5^{\max(1,0)}\cdot11^{\max(0,1)}
=2^{2}\cdot3^{3}\cdot5\cdot11=5940.
\]

Ratio:
\[
\frac{\text{lcm}}{\text{gcd}} = \frac{5940}{18}=330.
\]

ANSWER 3: C

---

Problem 4:
Starting with \$2, each day Taye either adds \$3 or doubles his amount.

List all sequences of three operations (A = add 3, D = double) and the resulting amounts:

\[
\begin{array}{c|c}
\text{Sequence} & \text{Final amount}\\ \hline
AAA & 11\\
AAD & 16\\
ADA & 13\\
ADD & 20\\
DAA & 10\\
DAD & 14\\
DDA & 11\\
DDD & 16
\end{array}
\]

Distinct amounts are \(\{10,11,13,14,16,20\}\) – six possibilities.

ANSWER 4: D

---

Problem 5:
Let the three equally spaced integers be \(a-d,\ a,\ a+d\).

\[
\begin{cases}
(a-d)+a = 40 \;\Longrightarrow\; 2a-d = 40\\
a+(a+d)=60 \;\Longrightarrow\; 2a+d = 60
\end{cases}
\]

Add the equations: \(4a = 100 \Rightarrow a = 25\).  
Then \(2a-d = 40 \Rightarrow 50-d = 40 \Rightarrow d = 10\).

The three numbers are \(15,25,35\); their sum is \(15+25+35 = 75\).

ANSWER 5: B

---

Problem 6:
Maximum amount occurs when as many coins as possible are nickels (5 c).  
Since at least one penny is required, use 2019 nickels and 1 penny:

\[
2019\cdot5 + 1 = 10096\text{ cents}.
\]

Minimum amount occurs when as many coins as possible are pennies.  
Use 2019 pennies and 1 nickel:

\[
2019\cdot1 + 5 = 2024\text{ cents}.
\]

Difference:
\[
10096 - 2024 = 8072\text{ cents}.
\]

ANSWER 6: C

---

Problem 7:
In an isosceles triangle at least two angles are equal. With angles \(70^\circ\) and \(x^\circ\):

1. \(70 = x \;\Rightarrow\; x = 70\).  
2. \(70\) equals the third angle \(\Rightarrow\; x = 40\).  
3. \(x\) equals the third angle \(\Rightarrow\; 70 + 2x = 180 \;\Rightarrow\; x = 55\).

The three possible values are \(40,55,70\); their sum is \(165^\circ\).

ANSWER 7: D

---

Problem 8:
\[
\frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2\cdot 3 = 6.
\]

ANSWER 8: E

---

Problem 9:
We need \(N \equiv 2 \pmod{3,4,5,6}\).  
Thus \(N-2\) is a multiple of \(\text{lcm}(3,4,5,6)=60\).  
The smallest \(N>2\) is \(N = 60+2 = 62\).

\(62\) lies between 60 and 79.

ANSWER 9: B

---

Problem 10:
Find ordered triples of digits \(a,b,c\) (each 1–9) with product 24.

Factor \(24 = 2^3\cdot3\). Only digits allowed are \(\{1,2,3,4,6,8\}\) (since 5,7,9 cannot appear).

Exactly one digit contributes a factor 3, so either a 3 or a 6.

- With a 3: remaining two digits must give \(2^3\). Pairs \((1,8)\) and \((2,4)\) → each yields 6 permutations → 12 numbers.
- With a 6: remaining two digits must give \(2^{2}\). Pairs \((1,4)\) (6 permutations) and \((2,2)\) (3 permutations) → 9 numbers.

Total \(12+9 = 21\) three‑digit numbers.

ANSWER 10: D

---

Problem 11:
Ben’s steps: start with 6, add 1 → 7, double → 14. He gives 14 to Sue.

Sue’s steps: subtract 1 → 13, double → 26.

ANSWER 11: C

---

Problem 12:
\(10001\) is odd. A sum of two primes is odd only when one prime is 2 and the other is odd.  

\(10001 - 2 = 9999\), which is composite \((9999 = 3^2\cdot11\cdot101)\).  

Thus no pair of primes adds to 10001.

ANSWER 12: A

Problem 1:  
The operation $a \ast b$ is defined as $3a - b$. We need to find $x$ such that $2 \ast (5 \ast x) = 1$.  
First, compute $5 \ast x$: $3 \cdot 5 - x = 15 - x$.  
Then, compute $2 \ast (15 - x)$: $3 \cdot 2 - (15 - x) = 6 - 15 + x = x - 9$.  
Set this equal to 1: $x - 9 = 1$, so $x = 10$.  
The choices include 10 (D), so other options are incorrect as they don't satisfy the equation.  
ANSWER 1: D  

Problem 2:  
Maria works 8 hours, not including a 45-minute lunch. She starts at 7:25 A.M. and takes lunch at noon.  
From 7:25 A.M. to 12:00 P.M. is 4 hours and 35 minutes. After lunch, she works the remaining time: 8 hours total - 4 hours 35 minutes = 3 hours 25 minutes.  
Adding this to 12:45 P.M. (noon + 45-minute lunch) gives 4:10 P.M.  
The choices include 4:10 P.M. (C), so other options are incorrect.  
ANSWER 2: C  

Problem 3:  
Find the ratio of LCM(180, 594) to GCF(180, 594).  
Prime factorizations: $180 = 2^2 \cdot 3^2 \cdot 5$, $594 = 2 \cdot 3^3 \cdot 11$.  
GCF: $2^1 \cdot 3^2 = 18$.  
LCM: $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$.  
Ratio: $5940 / 18 = 330$.  
The choices include 330 (C), so other options are incorrect.  
ANSWER 3: C  

Problem 4:  
Taye starts with \$2 on Monday. Each day, he either gains \$3 or doubles his money. We need the possible amounts on Thursday (3 days later).  
Day 1 (Mon): \$2  
Day 2 (Tue): \$2 + 3 = \$5 or \$2 \cdot 2 = \$4  
Day 3 (Wed):  
- From \$5: \$5 + 3 = \$8 or \$5 \cdot 2 = \$10  
- From \$4: \$4 + 3 = \$7 or \$4 \cdot 2 = \$8  
Day 4 (Thu):  
- From \$8: \$8 + 3 = \$11 or \$8 \cdot 2 = \$16  
- From \$10: \$10 + 3 = \$13 or \$10 \cdot 2 = \$20  
- From \$7: \$7 + 3 = \$10 or \$7 \cdot 2 = \$14  
Unique amounts: \$10, \$11, \$13, \$14, \$16, \$20 (6 values).  
The choices include 6 (D), so other options are incorrect.  
ANSWER 4: D  

Problem 5:  
Three equally spaced integers: let them be $a$, $a+d$, $a+2d$.  
Given $a + (a+d) = 40$ → $2a + d = 40$.  
$(a+d) + (a+2d) = 60$ → $2a + 3d = 60$.  
Subtract the first equation from the second: $(2a + 3d) - (2a + d) = 60 - 40$ → $2d = 20$ → $d = 10$.  
Substitute $d = 10$ into $2a + d = 40$: $2a + 10 = 40$ → $2a = 30$ → $a = 15$.  
The numbers are 15, 25, 35. Sum: $15 + 25 + 35 = 75$.  
The choices include 75 (B), so other options are incorrect.  
ANSWER 5: B  

Problem 6:  
Ricardo has 2020 coins: pennies (1 cent) and nickels (5 cents). Let $p$ be pennies, $n$ be nickels.  
$p + n = 2020$, $p \geq 1$, $n \geq 1$.  
Total money: $M = p + 5n$.  
Express $p = 2020 - n$, so $M = 2020 - n + 5n = 2020 + 4n$.  
Maximize/minimize $M$:  
- Max $M$: maximize $n$. $n \leq 2019$, so $M_{\text{max}} = 2020 + 4 \cdot 2019 = 2020 + 8076 = 10096$.  
- Min $M$: minimize $n$. $n \geq 1$, so $M_{\text{min}} = 2020 + 4 \cdot 1 = 2024$.  
Difference: $10096 - 2024 = 8072$.  
The choices include 8072 (C), so other options are incorrect.  
ANSWER 6: C  

Problem 7:  
An isosceles triangle has two angles of 70° and $x$°. Possible cases:  
1. Two angles are 70°, so $x = 180 - 2 \cdot 70 = 40$.  
2. One angle is 70°, and the other two are equal ($x$° each): $70 + 2x = 180$ → $2x = 110$ → $x = 55$.  
3. Two angles are $x$°, and one is 70°: $2x + 70 = 180$ → $2x = 110$ → $x = 55$ (same as case 2) or if $x$ is the vertex angle, but in isosceles, if base angles are equal, so if $x$ is the unique angle, then $2 \cdot 70 + x = 180$ → $x = 40$ (same as case 1). Actually, the distinct cases are: two 70° angles ($x=40$), or one 70° and two $x$° ($x=55$), or two $x$° and one 70° ($x=55$ same). But if $x$ is the vertex angle, then $x = 180 - 2 \cdot 70 = 40$ (same as first). Actually, the three possibilities are: $x$ could be 70 (if two angles are $x$), but the problem says "two angles" are given as 70 and $x$, so $x$ is not necessarily the third. Better: the triangle has angles 70, 70, $x$ or 70, $x$, $x$. So:  
- If angles are 70, 70, $x$: $x = 40$.  
- If angles are 70, $x$, $x$: $70 + 2x = 180$ → $x = 55$.  
- Also, if $x$ is 70, but that's already covered. The problem says "two angles", so $x$ could be the third angle. But in the isosceles triangle, either two angles are equal to 70, or two angles are equal to $x$. So the distinct $x$ values are 40 and 55? But 70 is also possible if $x=70$. If $x=70$, then angles are 70, 70, 70? But the problem says "two angles" are 70 and $x$, so if $x=70$, then it's two 70s and another 70. But the sum is 180, so yes. So $x$ can be 40, 55, or 70. Sum: $40 + 55 + 70 = 165$.  
The choices include 165 (D), so other options are incorrect.  
ANSWER 7: D  

Problem 8:  
Compute $\frac{2}{1 - \frac{2}{3}}$.  
First, $1 - \frac{2}{3} = \frac{1}{3}$.  
Then, $\frac{2}{\frac{1}{3}} = 2 \cdot 3 = 6$.  
The choices include 6 (E), so other options are incorrect.  
ANSWER 8: E  

Problem 9:  
A number leaves remainder 2 when divided by 3, 4, 5, 6. So $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{4}$, $N \equiv 2 \pmod{5}$, $N \equiv 2 \pmod{6}$.  
Since 3,4,5,6 are not pairwise coprime, find LCM of 3,4,5,6. LCM(3,4,5,6) = LCM(12,5,6) = LCM(12,30) = 60? 60 divisible by 3,4,5,6? 60/3=20, 60/4=15, 60/5=12, 60/6=10, yes. So $N = 60k + 2$ for integer $k \geq 1$.  
Smallest $N > 2$: $k=1$ → $N=62$.  
62 is between 60 and 79.  
The choices include 60 and 79 (B), so other options are incorrect.  
ANSWER 9: B  

Problem 10:  
Find 3-digit numbers with digit product 24.  
Factorize 24: $24 = 2^3 \cdot 3$.  
Possible digit combinations (each digit 1-9):  
- 3,8,1: permutations: 3! = 6 (381,318,831,813,138,183)  
- 4,6,1: permutations: 6  
- 2,6,2: permutations: $\frac{3!}{2!} = 3$ (262,622,226)  
- 2,3,4: permutations: 6  
Total: 6 + 6 + 3 + 6 = 21.  
The choices include 21 (D), so other options are incorrect.  
ANSWER 10: D  

Problem 11:  
Ben thinks of 6.  
Step 1: Add 1: $6 + 1 = 7$.  
Step 2: Double: $7 \cdot 2 = 14$. Ben gives 14 to Sue.  
Sue's steps:  
Step 1: Subtract 1: $14 - 1 = 13$.  
Step 2: Double: $13 \cdot 2 = 26$.  
The choices include 26 (C), so other options are incorrect.  
ANSWER 11: C  

### Problem 1:
We are given a custom operation \(a*b = 3a - b\), and need to solve \(2*(5*x) = 1\) for \(x\). First evaluate the inner operation per order of rules:
1. Calculate \(5*x = 3(5) - x = 15 - x\).
2. Substitute back into the full expression: \(2*(15-x) = 3(2) - (15 - x) = 6 -15 +x = x-9\).
3. Set equal to 1: \(x-9=1 \implies x=10\).
Tempting wrong answers come from misordering operations or misapplying the custom operation definition.
ANSWER 1: D

---

### Problem 2:
Maria works 8 hours total, plus a 45-minute unpaid lunch. She starts at 7:25 AM, takes lunch at noon. First calculate total time spent at work (including lunch): \(8h45m\). Add this to her start time:
1. 7:25 AM + 8 hours = 3:25 PM.
2. Add the 45-minute lunch: 3:25 PM + 45 minutes = 4:10 PM.
Common errors: Forgetting to add lunch time or miscalculating pre-lunch work hours.
ANSWER 2: C

---

### Problem 3:
We need the ratio \(\frac{\text{LCM}(180,594)}{\text{GCF}(180,594)}\). Use prime factorization first:
1. \(180 = 2^2 \times 3^2 \times 5\), \(594 = 2 \times 3^3 \times 11\).
2. GCF is the product of minimum exponents of shared primes: \(2^1 \times 3^2 = 18\).
3. LCM is the product of maximum exponents of all primes: \(2^2 \times 3^3 \times5 \times11 = 5940\).
4. Ratio: \(5940 / 18 = 330\).
Wrong answers come from miscalculating LCM or GCF values.
ANSWER 3: C

---

### Problem 4:
Taye starts with $2 on Monday, each day he adds $3 or doubles his money. We count unique amounts on Thursday (3 days later). There are \(2^3=8\) total operation sequences, leading to these unique results: 10,11,13,14,16,20. That is 6 distinct values. Duplicates (11 and 16 appear twice from different sequences) do not add to the count.
ANSWER 4: D

---

### Problem 5:
Three equally spaced integers, so call them \(n, n+d, n+2d\). We know:
1. \(n + (n+d) = 40 \implies 2n+d=40\)
2. \((n+d)+(n+2d)=60 \implies 2n+3d=60\)
Subtract the first equation from the second: \(2d=20 \implies d=10\), so \(n=15\). The three numbers are 15,25,35, sum = 75.
ANSWER 5: B

---

### Problem 6:
Ricardo has 2020 coins, at least one penny (1c) and one nickel (5c). Calculate max and min total value:
1. Max value: 1 penny, 2019 nickels = \(1 + 2019*5 = 10096\)c.
2. Min value: 2019 pennies, 1 nickel = \(2019 + 5 = 2024\)c.
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C

---

### Problem 7:
For an isosceles triangle with angles 70° and \(x^\circ\), the only valid values of \(x\) are:
1. \(x=40\): angles 70,70,40
2. \(x=55\): angles 70,55,55
3. \(x=70\): angles 70,70,40
Sum of the three values: \(40+55+70=165\).
ANSWER 7: D

---

### Problem 8:
Simplify the expression step-by-step:
1. Denominator: \(1 - 2/3 = 1/3\)
2. Full expression: \(2 / (1/3) = 6\)
ANSWER 8: E

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### Problem 9:
A number leaving remainder 2 when divided by 3,4,5,6 means the number minus 2 is divisible by all four values. The LCM of 3,4,5,6 is 60, so the smallest such number is \(60+2=62\), which lies between 60 and 79.
ANSWER 9: B

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### Problem 10:
Count 3-digit numbers with digit product 24 by listing all valid digit triples and their permutations:
1. (1,3,8): 6 permutations
2. (1,4,6): 6 permutations
3. (2,2,6): 3 permutations
4. (2,3,4):6 permutations
Total: 6+6+3+6=21.
ANSWER 10: D

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### Problem 11:
Ben thinks of 6: add 1 to get 7, double to get 14, which he gives to Sue. Sue subtracts 1 from 14 to get 13, double to get 26.
ANSWER 11: C

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### Problem 12:
10001 is odd, so it can only be written as the sum of an even and odd prime. The only even prime is 2, so the other number would be \(10001-2=9999\), which is composite (divisible by 9). There are no valid ways to write 10001 as the sum of two primes, so the number of ways is 0.
ANSWER 12: A

### Problem 1:
The operation \(a * b\) is defined as \(3a - b\). We need to solve \(2 * (5 * x) = 1\). Operations are evaluated innermost first:
1.  Compute \(5 * x\): \(3(5) - x = 15 - x\).
2.  Substitute into the outer operation: \(2 * (15 - x) = 3(2) - (15 - x) = 6 - 15 + x = x - 9\).
3.  Set equal to 1: \(x - 9 = 1 \implies x = 10\).
ANSWER 1: D
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### Problem 2:
Maria works 8 hours total, plus a 45-minute lunch break, so total elapsed time is \(8\text{h}45\text{m}\). She starts at 7:25 AM:
1.  Add 8 hours to 7:25 AM: \(7:25 + 8\text{h} = 3:25\text{ PM}\).
2.  Add 45 minutes: \(3:25 + 45\text{m} = 4:10\text{ PM}\).
ANSWER 2: C
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### Problem 3:
For any two positive integers, \(\text{LCM}(a,b) \times \text{GCF}(a,b) = a \times b\), so \(\frac{\text{LCM}(a,b)}{\text{GCF}(a,b)} = \frac{a \times b}{(\text{GCF}(a,b))^2}\). First factorize 180 and 594:
- \(180 = 2^2 \times 3^2 \times 5\)
- \(594 = 2 \times 3^3 \times 11\)
GCF is the product of the smallest exponents of common primes: \(2^1 \times 3^2 = 18\).
LCM is the product of the largest exponents of all primes: \(2^2 \times 3^3 \times 5 \times 11 = 5940\).
Ratio: \(\frac{5940}{18} = 330\).
ANSWER 3: C
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### Problem 4:
Taye starts with $2, and over 3 days (Monday to Thursday) chooses to either add $3 or double the current amount each day. We list all 8 possible sequences of choices, then count unique results:
1.  Add, Add, Add: \(2 \to 5 \to 8 \to 11\)
2.  Add, Add, Double: \(2 \to5 \to8 \to16\)
3.  Add, Double, Add: \(2 \to5 \to10 \to13\)
4.  Add, Double, Double: \(2 \to5 \to10 \to20\)
5.  Double, Add, Add: \(2 \to4 \to7 \to10\)
6.  Double, Add, Double: \(2 \to4 \to7 \to14\)
7.  Double, Double, Add: \(2 \to4 \to8 \to11\) (duplicate)
8.  Double, Double, Double: \(2 \to4 \to8 \to16\) (duplicate)
Unique amounts: 10, 11, 13, 14, 16, 20 → 6 total.
ANSWER 4: D
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### Problem 5:
Three equally spaced integers form an arithmetic sequence: let them be \(a-d, a, a+d\) where \(d\) is the common difference.
1.  Sum of first and second: \((a-d) + a = 2a - d = 40\)
2.  Sum of second and third: \(a + (a+d) = 2a + d = 60\)
3.  Add the two equations: \(4a = 100 \implies a=25\). Substitute back to find \(d=10\).
4.  Sum of all three: \(3a = 75\).
ANSWER 5: B
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### Problem 6:
Let \(p\) = number of pennies, \(n\) = number of nickels. Then \(p + n = 2020\), and total value \(V = p + 5n\). Substitute \(p = 2020 - n\):
\(V = (2020 - n) + 5n = 2020 + 4n\).
To maximize \(V\), maximize \(n\): at least 1 penny, so \(n \leq 2019\). Max \(V = 2020 + 4(2019) = 10096\).
To minimize \(V\), minimize \(n\): at least 1 nickel, so \(n \geq1\). Min \(V = 2020 +4(1) = 2024\).
Difference: \(10096 - 2024 = 8072\).
ANSWER 6: C
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### Problem 7:
An isosceles triangle has at least two equal angles, and total angle sum is 180°. We know one angle is 70°, another is \(x°\), so the third angle is \(180 -70 -x = 110 -x\). For the triangle to be isosceles, at least two angles are equal, giving three cases:
1.  \(70 = x\): Then the third angle is \(110 -70 =40\), valid (angles 70,70,40).
2.  \(70 = 110 -x\): Solve to get \(x=40\), valid (angles 70,70,40).
3.  \(x = 110 -x\): Solve to get \(2x=110 \implies x=55\), valid (angles70,55,55).
The three possible \(x\) values are 40, 55, 70. Sum: \(40+55+70=165\).
ANSWER 7: D
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### Problem 8:
Compute the denominator first: \(1 - \frac{2}{3} = \frac{1}{3}\). Then divide: \(\frac{2}{\frac{1}{3}} = 2 \times 3 =6\).
ANSWER 8: E
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### Problem 9:
If a number \(N\) leaves remainder 2 when divided by 3,4,5,6, then \(N-2\) is divisible by all these numbers. The smallest such \(N>2\) uses the least common multiple (LCM) of 3,4,5,6:
- LCM of 3,4,5,6 is \(2^2 \times3 \times5=60\).
- So \(N-2=60 \implies N=62\).
62 lies between 60 and 79.
ANSWER 9: B
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### Problem 10:
We need 3-digit numbers (hundreds digit \(a \geq1\), tens \(b\geq0\), units \(c\geq0\)) with \(a \times b \times c=24\). Since 24 is non-zero, all digits are at least 1. Factorize 24 into triples of digits (1-9):
1.  (1,3,8): all distinct, \(3! =6\) permutations.
2.  (1,4,6): all distinct, \(3! =6\) permutations.
3.  (2,2,6): two identical, \(3\) distinct permutations.
4.  (2,3,4): all distinct, \(3! =6\) permutations.
Total: \(6+6+3+6=21\).
ANSWER 10: D
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### Problem 11:
Ben thinks of 6:
1.  Add 1: \(6+1=7\), double: \(7 \times2=14\). He gives 14 to Sue.
2.  Sue subtracts 1: \(14-1=13\), doubles: \(13 \times2=26\).
ANSWER 11: C
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### Problem 12:
10001 is odd. The sum of two primes is odd only if one prime is even (the only even prime is 2) and the other is odd. So the only possible pair is \(2 + (10001 -2) = 2 + 9999\). Check if 9999 is prime: sum of digits is \(9+9+9+9=36\), divisible by 3, so 9999 is composite. No valid pairs exist.
ANSWER 12: A